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IN   MEMORIAM 
FLORIAN  CAJORl 


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WENTWORTH'S 
SOLID  GEOMETRY 


REVISED  BY 

GEORGE  WENTWORTH 

AND 

DAVID  EUGENE  SMITH 


GINN  AND  COMPANY 

BOSTON  •  NEW   YORK  •  CHICAGO  •  LONDON 


COPYRIGHT,  1888,  1899,  BY  G.  A.  WENTWORTH 


COPYRIGHT,  1911,  BY  GEORGE  WENTWORTH 

AND   DAVID   EUGENE   SMITH 

ENTERED   AT  STATIONERS'  HALL 

ALL  BIGHTS  RESERVED 

911.8 


GINN  AND  COMPANY  •  PRO- 
PRIETORS •  BOSTON  •  U.S.A. 


PREFACE 


Long  after  the  death  of  Eobert  Eecorde,  England's  first 
great  writer  of  textbooks,  the  preface  of  a  new  edition  of 
one  of  his  works  contained  the  appreciative  statement  that 
the  book  was  "entail'd  upon  the  People,  ratified  and  signed 
by  the  approbation  of  Time."  The  language  of  this  sentiment 
sounds  quaint,  but  the  noble  tribute  is  as  impressive  to-day 
as  when  first  put  in  print  two  hundred  fifty  years  ago. 

With  equal  truth  these  words  may  be  applied  to  the  Geom- 
etry written  by  George  A.  Wentworth.  For  a  generation  it 
has  been  the  leading  textbook  on  the  subject  in  America.  It 
set  a  standard  for  usability  that  every  subsequent  writer  upon 
geometry  has  tried  to  follow,  and  the  number  of  pupils  who 
have  testified  to  its  excellence  has  run  well  into  the  millions. 

In  undertaking  to  prepare  a  revision  of  the  Solid  Geometry, 
the  authors  have  been  guided  by  certain  well-defined  princi- 
ples, based  upon  an  extended  investigation  of  the  needs  of  the 
schools  and  upon  a  study  of  all  that  is  best  in  the  recent  liter- 
ature of  the  subject.  The  effects  of  these  principles  they  feel 
should  be  summarized  for  the  purpose  of  calling  the  attention 
of  the  wide  circle  of  friends  of  the  Wentworth  text  to  the 
points  of  similarity  and  of  difference  in  the  editions. 

1.  Every  effort  has  been  made  not  only  to  preserve  but  to 
improve  upon  the  simplicity  of  treatment,  the  clearness  of  ex- 
pression, and  the  symmetry  of  page  that  have  characterized 
the  successive  editions  of  the  Wentworth  Geometry.  It  has 
been  the  purpose  to  prepare  a  book  that  should  do  even  more 
than  maintain  the  traditions  this  work  has  fostered. 


•m^^^^^^.M  ^-^  M 


iv  SOLID  GEOMETRY 

2.  The  proofs  have  been  given  substantially  in  full,  to  the 
end  that  the  pupil  may  always  have  before  him  a  model  for 
his  independent  treatment  of  the  exercises. 

3.  To  meet  a  general  demand,  the  number  of  propositions 
has  been  decreased  so  as  to  include  only  the  great  basal  theo- 
rems and  problems.  A  little  of  the  less  important  material 
has  been  placed  in  the  Appendix,  to  be  used  or  not  as  circum- 
stances demand. 

4.  The  exercises,  in  some  respects  the  most  important  part 
of  a  course  in  geometry,  have  been  rendered  more  dignified  in 
appearance  and  have  been  improved  in  content.  The  number 
of  simple  exercises  has  been  greatly  increased,  while  the  diffi- 
cult puzzle  is  much  less  in  evidence  than  in  most  American 
textbooks.  The  exercises  are  systematically  grouped,  appear- 
ing in  general  in  full  pages,  in  large  type,  and  at  frequent 
intervals.  They  are  not  all  intended  for  one  class,  but  are  so 
numerous  as  to  allow  the  teacher  to  make  selections  from  year 
to  year. 

5.  The  work  throughout  has  been  made  as  concrete  as  is 
reasonable.  Definitions  have  been  postponed  until  they  are 
actually  needed,  only  well-recognized  terms  have  been  em- 
ployed, the  pupil  is  led  to  apply  his  geometry  to  practical  cases 
in  mensuration,  and  correlation  is  made  with  the  algebra 
already  studied. 

6.  All  the  references  to  Plane  Geometry  that  are  directly 
made  in  the  proof  of  Solid  Geometry  have  been  prefixed  to  this 
edition  so  as  to  be  easily  accessible. 

The  authors  are  indebted  to  many  friends  of  the  Went  worth 
Geometry  for  assistance  and  encouragement  in  the  labor  of 
preparing  this  edition,  and  they  will  welcome  any  further  sug- 
gestions for  improvement  from  any  of  their  readers. 

,     GEORGE  WENTWORTH 
DAVID  EUGENE  SMITH 


CONTENTS 


Pagk 

REFERENCES  TO  PLANE  GEOMETRY    .         .         .         ,         .  vii 

BOOK  VI.    LINES  AND  PLANES  IN  SPACE    .         .         .         .273 

Lines  and  Planes 273 

Dihedral  Angles    .........  293 

Polyhedral  Angles         ......"..  308 

Exercises 314 

BOOK  VII.    POLYHEDRONS,  CYLINDERS,  AND  CONES       .  317 

Polyhedrons 317 

Prisms 317 

Parallelepipeds 322 

Pyramids  ...........  337 

Regular  Polyhedrons 350 

Cylinders 353 

Cones 362 

Exercises 376 

BOOK  VIII.    THE  SPHERE  .         .         .         .         .         .         .381 

Spheres 381 

Plane  Sections  and  Tangent  Planes    .....  382 

Spherical  Polygons 392 

Measurement  of  Spherical  Surfaces  .....  410 

Measurement  of  Spherical  Solids       .....  421 

Exercises 424 

APPENDIX 431 

Polyhedrons    .         .         .         .         .         .         .         .         .         .  432 

Spherical  Segments         ........  444 

Recreations  of  Geometry     .......  449 

History  of  Geometry 453 

INDEX             459 

V 


SYMBOLS  AND  ABBREVIATIONS 


=   equals,  equal,  equal  to. 

Adj. 

adjacent. 

is  equal  to,  oi- 

Alt. 

alternate. 

ls  equivalent  to. 

Ax. 

axiom. 

>  is  greater  than. 

Const. 

construction^   . 

<  is  less  than. 

Cor. 

corollary. 

II  parallel. 

Def. 

definition. 

_L  perpendicular. 

Ex. 

exercise. 

Z  angle. 

Ext. 

exterior. 

A  triangle. 

Eig. 

figure. 

O  parallelogram. 

Hyp. 

hypothesis. 

□  rectangle. 

Iden. 

identity. 

O  circle. 

Int. 

interior. 

st.  straight. 

Post. 

postulate. 

rt.  right. 

Prob. 

problem. 

■.•  since. 

Prop. 

proposition. 

.*.  therefore. 

Sup. 

supplementary. 

These  symbols  take  the  phiral  form  when  necessary,  as  in  the  case  of 
lis,  A,  A,  (D. 

The  symbols  +,  — ,  x  ,  -4-  are  used  as  in  algebra. 

There  is  no  generally  accepted  symbol  for  "is  congruent  to,"  and  the 
words  are  used  in  this  book.  Some  teachers  use  =  or  =,  and  some  use 
=  ,  but  the  sign  of  equality  is  more  commonly  employed,  the  context 
telling  whether  equality,  equivalence,  or  congruence  is  to  be  understood. 

Q.  E.D.  is  an  abbreviation  that  has  long  been  used  in  geometry  for 
the  Latin  words  quod  erat  demonstrandum,  ''which  was  to  be  proved." 

Q.  E.  F.  stands  for  quod  erat  faciendum,  ''  which  was  to  be  done." 


VI 


REFERENCES  TO  PLANE  CxEOMETRY 

28.  A  portion  of  a  plane  bounded  by  three  straight  lines  is 
called  a  triangle. 

41.  The  whole  angular  space  in  a  plane  about  a  point  is 
called  a  perigon. 

52.  The  following  are  the  most  important  axioms  used  in 
geometry : 

1.  If  equals  are  added  to  equals,  the  sums  are  equal. 

2.  If  equals  are  subtracted  from  equals,  the  remainders  are 
equal. 

3.  If  equals  are  multiplied  by  equals,  the  products  are  equal. 

4.  If  equals  are  divided  by  equals,  the  quotients  are  equal. 
In  division  the  divisor  is  never  zero. 

5.  Like  powers  and  like  positive  roots  of  equals  are  equal. 

6.  If  unequals.are  operated  on  by  positive  equals  in  the 
same  way^  the  results  are  unequal  in  the  same  order. 

7.  If  unequals  are  added  to  unequals  in  the  same  order,  the 
sums  are  unequal  in  the  same  order  ;  if  unequals  are  subtracted 
from  equals,  the  remainders  are  unequal  in  the  reverse  order. 

8.  Quantities  that  are  equal  to  the  same  quantity  or  to  equal 
quantities  are  equal  to  each  other. 

9.  A  quantity  may  be  substituted  for  its  equal  in  an  equa- 
tion or  in  an  inequality. 

10.  If  the  first  of  three  quantities  is  greater  than  the  second, 
and  the  second  is  greater  than  the  third,  then  the  first  is  greater 
than  the  third. 

11.  The  whole  is  greater  than  any  of  its  parts,  and  is  equal 
to  the  sum  of  all  its  parts. 


viii  SOLID  GEOMETRY 

53.  Postulate  5.  Any  figure  may  be  moved  from  one  place 
to  another  without  altering  its  size  or  shape. 

56.  All  right  angles  are  equal. 

57.  From  a  given  point  in  a  given  line  only  one  perpendic- 
ular can  be  drawn  to  the  line. 

60.  If  two  lines  intersect,  the  vertical  angles  are  equal. 

67.  Corresponding  parts  of  congruent  figures  are  equal. 

68.  Two  triangles  are  congruent,  if  two  sides  and  the  included 
angle  of  the  one  are  equal  respectively  to  two  sides  and  the 
included  angle  of  the  other. 

69.  Two  right  triangles  are  congruent,  if  the  sides  of  the 
right  angles  are  equal  respectively. 

72.  Two  triangles  are  congruent,  if  twD  angles  and  the  in- 
cluded side  of  the  one  are  equal  respectively  to  two  angles  and 
the  included  side  of  the  other. 

80.  Two  triangles  are  congruent,  if  the  three  sides  of  the  one 
are  equal  respectively  to  the  three  sides  of  the  other. 

82.  Only  one  perpendicular  can  be  drawn  to  a  given  line 
from  a  given  external  point. 

84.  Of  two  lines  drawn  from  a  point  in  a  perpendicular  to  a 
given  line,  cutting  off  on  the  given  line  unequal  segments  from 
the  foot  of  the  perpendicular,  the  more  remote  is  the  greater. 

89.  Two  right  triangles  are  congruent,  if  the  hypotenuse  and 
a  side  of  the  one  are  equal  respectively  to  the  hypotenuse 
and  a  side  of  the  other. 

93.  Lines  that  lie  in  the  same  plane  and  cannot  meet  how- 
ever far  produced  are  called  parallel  lines. 

94.  Through  a  given  point  only  one  line  can  be  drawn  par- 
allel to  a  given  line. 

95.  Two  lines  in  the  same  plane  perpendicular  to  the  same 
line  are  parallel. 


REFERENCES  TO  PLANE  GEOMETRY  ix 

97.  If  a  line  is  perpendicular  to  one  of  two  parallel  lines,  it 
is  perpendicular  to  the  other  also. 

112.  The  sum  of  any  two  sides  of  a  triangle  is  greater  than 
the  third  side,  and  the  difference  between  any  two  sides  is 
less  than  the  third  side. 

116.  If  two  triangles  have  two  sides  of  the  one  equal  respec- 
tively to  two  sides  of  the  other,  but  the  third  side  of  the  first 
triangle  greater  than  the  third  side  of  the  second,  then  the 
angle  opposite  the  third  side  of  the  first  is  greater  than  the 
angle  opposite  the  third  side  of  the  second.  - 

118.  A  quadrilateral  may  be  a  trapezoid,  having  two  sides 
parallel;  a  parallelogram,  having  the  opposite  sides  parallel; 
or  it  may  have  no  sides  parallel. 

125.  The  opposite  sides  of  a  parallelogram  are  equal. 

126.  A  diagonal  divides  a  parallelogram  into  two  congruent 
triangles. 

127.  Segments  of  parallel  lines  cut  off  by  parallel  lines  are 
equal. 

130.  If  two  sides  of  a  quadrilateral  are  equal  and  parallel, 
then  the  other  two  sides  are  equal  and  parallel,  and  the  figure 
is  a  parallelogram. 

131.  The  diagonals  of  a  parallelogram  bisect  each  other. 

132.  Two  parallelograms  are  congruent,  if  two  sides  and  the 
included  angle  of  the  one  are  equal  respectively  to  two  sides 
and  the  included  angle  of  the  other. 

133.  Two  rectangles  having  equal  bases  and  equal  altitudes 
are  congruent. 

136.  The  line  which  joins  the  mid-points  of  two  sides  of  a 
triangle  is  parallel  to  the  third  side,  and  is  equal  to  half  the 
third  side. 


X  SOLID  GEOMETKY 

142.  Two  polygons  are 

mutually  equiangular,  if  the  angles  of  the  one  are  equal  to 
the  angles  of  the  other  respectively,  taken  in  the  same  order ; 

mutually  equilateral,  if  the  sides  of  the  one  are  equal  to  the 
sides  of  the  other  respectively,  taken  in  the  same  order ; 

congruent,  if  mutually  equiangular  and  mutually  equilateral, 
since  they  then  can  be  made  to  coincide. 

145.  Each  angle  of  a  regular  polygon  of  ?i  sides  is  equal  to 

— ^^ right  angles. 

n 

146.  The  sum  of  the  exterior  angles  of  a  polygon,  made  by 
producing  each  of  its  sides  in  succession,  is  equal  to  four  right 
angles. 

148.  To  prove  that  a  certain  line  or  group  of  lines  is  the 
locus  of  a  point  that  fulfills  a  given  condition,  it  is  necessary 
and  sufficient  to  prove  two  things  : 

1.  That  any  point  in  the  supposed  locus  satisfies  the  con- 
dition. 

2.  That  any  point  outside  the  supposed  locus  does  not  satisfy 
the  given  condition. 

150.  The  locus  of  a  point  equidistant  from  the  extremities 
of  a  given  line  is  the  perpendicular  bisector  of  that  line. 

151.  Two  points  each  equidistant  from  the  extremities  of  a 
line  determine  the  perpendicular  bisector  of  the  line. 

152.  The  locus  of  a  point  equidistant  from  two  given  inter- 
secting lines  is  a  pair  of  lines  bisecting  the  angles  formed  by 
those  lines. 

159.  A  closed  curve  lying  in  a  plane,  and  such  that  all  of 
its  points  are  equally  distant  from  a  fixed  point  in  the  plane, 
is  called  a  circle. 


EEFERENCES  TO  PLANE  GEOMETRY  xi 

162.  All  radii  of  the  same  circle  or  of  equal  circles  are  equal ; 
and  all  circles  of  equal  radii  are  equal. 

167.  In  the  same  circle  or  in  equal  circles  equal  arcs  subtend 
equal  central  angles ;  and  of  two  unequal  arcs  the  greater  sub- 
tends the  greater  central  angle. 

172.  In  the  same  circle  or  in  equal  circles,  if  two  chords  are 
equal,  they  subtend  equal  arcs  ;  and  if  two  chords  are  unequal, 
the  greater  subtends  the  greater  arc 

174.  A  line  through  the  center  of  a  circle  perpendicular  to 
a  chord  bisects  the  chord  and  the  arcs  subtended  by  it. 

178.  In  the  same  circle  or  in  equal  circles  equal  chords  are 
equidistant  from  the  center,  and  chords  equidistant  from  the 
center  are  equal. 

185.  A  tangent  to  a  circle  is  perpendicular  to  the  radius 
drawn  to  the  point  of  contact. 

195.  If  two  circles  intersect,  the  line  of  centers  is  the  per- 
pendicular bisector  of  their  common  chord. 

204.  When  a  variable  approaches  a  constant  in  such  a  way 
that  the  diiference  between  the  two  may  become  and  remain 
less  than  any  assigned  positive  quantity,  however  small,  the 
constant  is  called  the  limit  of  the  variable. 

207.  If,  while  approaching  their  respective  limits,  two  vari- 
ables are  always  equal,  their  limits  are  equal. 

212.  In  the  same  circle  or  in  equal  circles  two  central  angles 
have  the  same  ratio  as  their  intercepted  arcs. 

213.  A  central  angle  is  measured  by  the  intercepted  arc. 

261.  In  any  proportion  the  product  of  the  extremes  is  equal 
to  the  product  of  the  means. 

262.  The  mean  proportional  between  two  quantities  is  equal 
to  the  square  root  of  their  product. 


xii  SOLID  GEOMETRY 

269.  In  a  series  of  equal  ratios,  the  sum  of  the  antecedents 
is  to  the  sum  of  the  consequents  as  any  antecedent  is  to  its 
consequent. 

270.  Like  powers  of  the  terms  of  a  proportion  are  in  pro- 
portion. 

273.  If  a  line  is  drawn  through  two  sides  of  a  triangle  par- 
allel to  the  third  side,  it  divides  the  two  sides  proportionally. 

274.  One  side  of  a  triangle  is  to  either  of  its  segments  cut 
off  by  a  line  parallel  to  the  base  as  the  third  side  is  to  its 
corresponding  segment. 

275.  Three  or  more  parallel  lines  cut  off  proportional  in- 
tercepts on  any  two  transversals. 

282.  Polygons  that  have  their  corresponding  angles  equal, 
and  their  corresponding  sides  proportional,  are  called  similar 
polygons. 

285.   Two  mutually  equiangular  triangles  are  similar. 

288.  If  two  triangles  have  an  angle  of  the  one  equal  to  an 
angle  of  the  other,  and  the  including  sides  proportional,  they 
are  similar. 

289.  If  two  triangles  have  their  sides  respectively  propor- 
tional, they  are  similar. 

290.  Two  triangles  which  have  their  sides  respectively  par- 
allel, or  respectively  perpendicular,  are  similar. 

292.  If  two  polygons  are  similar,  they  can  be  separated 
into  the  same  number  of  triangles,  similar  each  to  each,  and 
similarly  placed. 

298.  If  a  perpendicular  is  drawn  from  any  point  on  a  circle 
to  a  diameter,  the  chord  from  that  point  to  either  extremity  of 
the  diameter  is  the  mean  proportional  between  the  diameter 
and  the  segment  adjacent  to  that  chord. 


EEFEREKCES  TO  PLANE  GEOMETRY         xiii 

322.  The  area  of  a  parallelogram  is  equal  to  the  product  of 
its  base  by  its  altitude. 

323.  Parallelograms  having  equal  bases  and  equal  altitudes 
are  equivalent. 

325.  The  area  of  a  triangle  is  equal  to  half  the  product  of 
its  base  by  its  altitude. 

326.  Triangles  having  equal  bases  and  equal  altitudes  are 
equivalent. 

327.  Triangles  having  equal  bases  are  to  each  other  as  their 
altitudes  ;  triangles  having  equal  altitudes  are  to  each  other  as 
their  bases ;  any  two  triangles  are  to  each  other  as  the  prod- 
ucts of  their  bases  by  their  altitudes. 

329.  The  area  of  a  trapezoid  is  equal  to  half  the  product  of 
the  sum  of  its  bases  by  its  altitude. 

332.  The  areas  of  two  triangles  that  have  an  angle  of  the 
one  equal  to  an  angle  of  the  other  are  to  each  other  as  the 
products  of  the  sides  including  the  equal  angles. 

334.  The  areas  of  two  similar  polygons  are  to  each  other  as 
the  squares  on  any  two  corresponding  sides. 

377.  If  the  number  of  sides  of  a  regular  inscribed  polygon 
is  indefinitely  increased,  the  apothem  of  the  polygon  approaches 
the  radius  of  the  circle  as  its  limit. 

381.  The  circle  is  the  limit  which  the  perimeters  of  regular 
inscribed  polygons  and  of  similar  circumscribed  polygons  ap- 
proach, if  the  number  of  sides  of  the  polygons  is  indefinitely 
increased. 

The  area  of  the  circle  is  the  limit  which  the  areas  of  the 
inscribed  and  circumscribed  polygons  approach. 

.  382.  Two  circumferences  have  the  same  ratio  as  their  radii. 

385.  The  circumference  of  a  circle  equals  2  irr. 

389.  The  area  of  a  circle  =  irr^. 


SOLID  GEOMETRY 

BOOK  VI 

LINES  AND  PLANES  IN  SPACE 

421.  The  Nature  of  Solid  Geometry.  In  plane  geometry  we 
deal  with  figures  lying  in  a  flat  surface,  studying  their  proper- 
ties and  relations  and  measuring  the  figures.  In  solid  geometry 
we  shall  deal  with  figures  not  only  of  two  dimensions  lout  of 
three  dimensions,  also  studying  their  properties  and  relations 
and  measuring  the  figures. 

422.  Plane.  A  surface  such  that  a  straight  line  joining  any 
two  of  its  points  lies  wholly  in  the  surface  is  called  a  2jl(^ne. 

A  plane  is  understood  to  be  indefinite  in  extent,  but  it  is  conveniently 
represented  by  a  rectangle  seen  obliquely,  as  here  shown. 


423.  Determining  a  Plane.  A  plane  is  said  to  be  determined 
by  certain  lines  or  points  if  it  contains  the  given  lines  or 
points,  and  no  other  plane  can  contain  them. 

When  we  suppose  a  plane  to  be  drawn  to  include  given  points  or  lines, 
we  are  said  to  pass  the  plane  through  these  points  or  lines. 

When  a  straight  line  is  drawn  from  an  external  point  to  a  plane,  its 
point  of  contact  with  the  plane  is  called  its  foot. 

424.  Intersection  of  Planes.  The  line  that  contains  all  the 
points  common  to  two  planes  is  called  their  intersection. 

273 


274  BOOK  VI.    SOLID  GEOMETRY 

425.  Postulate  of  Planes.  Corresponding  to  the  postulate  that 
one  straight  line,  and  only  one,  can  be  drawn  through  two  given 
points,  the  following  postulate  is  assumed  for  planes : 

One  plane,  and  only  one,  can  he  passed  through  two  given 

intersecting  straight  lines. 

For  it  is  apparent  from  the  first  figure  that  a  plane  may  be  made  to 
turn  about  any  single  straight  line  AB,  thus  assuming  different  positions. 
But  if  CD  intersects  AB  at  P,  as  in  the  second  figure,  then  when  the 
plane  through  AB  turns  until  it  includes  C,  it  must  include  D,  since  it 
includes  two  points,  C  and  P,  of  the  line  (§  422) .  If  it  turns  any  more,  it 
will  no  longer  contain  C. 


426.  Corollary  1.    A  straight  line  and  a  point  not  in  the 
line  determine  a  plane. 

For  example,  line  AB  and  point  C  in  the  above  figure. 

427.  Corollary  2.    Three  points  not  in  a  straight  line  deter- 
mine a  plane. 

For  by  joining  any  one  of  them  with  the  other  two  we  have  two  inter- 
secting lines  (§  425). 

428.  Corollary  3.    Two  parallel  lines  determine  a  plane. 

M 

^  p=^         \ 

A B  \ 


N 
For  two  parallel  lines  lie  in  a  plane  (§  93),  and  a  plane  containinc 
either  parallel  and  a  point  P  in  the  other  is  determined  (§  426) . 


LINES  AND  PLANES 


275 


Proposition  I.    Theorem 

429.  If  two  jjlanes  cut  each  other,  their  intersection  is 
a  straight  line. 

p  .       N 


Given  ikfiV  and  PQy  two  planes  which  cut  each  other. 

To  prove  that  the  planes  MN  and  PQ  intersect  in  a 
straight   line. 

Proof.    Let  A  and  B  be  two  points  common  to  the  two  planes. 
Draw  a  straight  line  through  the  points  A  and  B. 
Then  the  straight  line  AB  lies  in  both  planes.      §  422 
{For  it  has  two  points  in  each  plane.) 

No  point  not  in  the  line  AB  can  be  in  both  planes ;  for  one 
plane,  and  only  one,  can  contain  a  straight  line  and  a  point 
without  the  line.  §  426 

Therefore  the  straight  line  through  A  and  B  contains  all 
the  points  common  to  the  two  planes,  and  is  consequently  the 
intersection  of  the  planes,  by  §  424.  q.e.d. 

Discussion.  What  is  the  corresponding  statement  in  plane  geometry  ? 

430.  Perpendicular  to  a  Plane.  If  a  straight  line  drawn  to  a 
plane  is  perpendicular  to  every  straight  line  that  passes  through 
its  foot  and  lies  in  the  plane,  it  is  said  to  be  per2:)endicular  to 
the  plane. 

When  a  line  is  perpendicular  to  a  plane,  the  plane  is  also  said  to  be 
perpendicular  to  the  line. 


276 


BOOK  VI.    SOLID  GEOMETRY 


Proposition  II.    Theorem 


431.  If  a  line  is  perpendicular  to  each  of  tivo  other 
lines  at  their  point  of  intersection,  it  is  perpendicular 
to  the  plane  of  the  tivo  lines. 


Given  the  line  AO  perpendicular  to  the  lines  OP  and  OR  at  O. 

To  prove  that  AO  is  1.  to  the  plane  MN  of  these  lines. 

Proof.  Through  0  draw  in  MN  any  other  line  OQ,  and  draw 
PR  cutting  OP,  OQ,  OR,  at  P,  Q,  and  R. 

Produce  AO  to  A',  making  OA'  equal  to  OA,  and  join  A  and 
A'  to  each  of  the  points  P,  Q,  and  R. 

Then  OP  and  OR  are  each  J_  to  AA^  at  its  mid-point. 

.'.AP  =  A'P,  ^ndAR  =  A'R.  §150 

.-.A  APR  is  congruent  to  A  A'PR.  §  80 

.•.ZRPA=ZA'PR.  §67 

That  is,  Z  QPA  =  ZA  'PQ. 

.'.  APQA  is  congruent  to  A  PQA\  §  68 

.    .•..4Q  =  ^'Q(§67);  and  6»Qis  ±  to.4.1'at  0.       §151 
.'.  AO  is  _L  to  any  and  hence  to  every  line  in  MN  through  0. 
.-.  .4  0  is  X  to  the  plane  MN,  by  §  430.  q.e.d 


LINES  AND  PLANES 
Proposition  III.    Theorem 


277 


432.  All  the  j^erpendiculars  that  can  he  drawn  to  a 
given  line  at  a  given  point  lie  in  a  plane  which  is  per- 
pendicular to  the  given  line  at  the  given  point. 


M 


R 


p- 


N 
Given  the  plane  MN  perpendicular  to  the  line  OY  at  0. 

To  prove  that  OF,  any  line  ±to  OY  at  0,  lies  in  MN. 

Proof.    Let  the  plane  containing  OY  and  OP  intersect  the 
plane  MN  in  the  line  OP';  then  OF  is  J_  to  OP'.  §  430 

In  the  plane  POY  only  one  ±  can  be  drawn  to  OF  at  0.   §  57 
Therefore  OP  and  OP'  coincide,  and  OP  lies  in  MN. 
Hence  every  ±  to  OF  at  O,  as  OQ,  OR,  lies  in  MN.         q.e.d. 

433.  Corollary  1.   Through  a  given  point  in  a  given  liyie 
one  plane,  and  only  one,  can  he  passed  perpendicular  to  the  line. 

434.  Corollary  2.  Through  a  given  external  point  one  plane, 
and  only  one,  can  he  passed  perpendicular  to  a  given  line. 

Given  the  line  OY  and  the  point  P.  F 

Draw  PO  ±  to  OY,  and  OQ  ±  to  OY. 
Then  OQ  and  OP  determine  a  plane  through 
P  ±  to  OY. 
.  Only  one  such  plane  can  be  drawn;  for 


M^ 


N 


only  one  ±  can  be  drawn  to  OF  from  the  point  P  (§  82). 

435.  Oblique  Line.   A  line  that  meets  a  plane  but  is  not  per- 
pendicular to  it  is  said  to  be  ohlique  to  the  plane. 


278 


BOOK  VI.    SOLID  GEOMETRY 
Proposition  IV.    Theorem 


436.  Through  a  given  point  in  a  plane  there  can  he 
drawn  one  line  perpendicular  to  the  plane,  and  only  one. 


Given  the  point  P  in  the  plane  MN. 


To  prove  that  there  can  he  drawn  one  line  perpendicular  to 
the  plane  MN  at  P,  and  only  one. 

Proof.  Through  the  point  P  draw  in  the  plane  MN  any  line 
AB,  and  pass  through  P  a  plane  XY  1.  to  AB,  cutting  the 
plane  MN  in  CD.  §  433 

At  P  erect  in  the  plane  XY  the  line  PQ  ±  to  CD. 
The  line  AB,  being  ±  to  the  plane  ZF  by  construction,  is  _L 
to  PQ,  which  passes  through  its  foot  in  the  plane.  §  430 

That  is,  PQ  is  J_  to  AB-,  and  as  it  is  ±  to  CD  by  construc- 
tion, it  is  ±  to  the  plane  MN.  §  431 
Moreover,  any  other  line  PR  drawn  from  P  is  oblique  to 
MN.    For  PQ  and  PR  intersecting  in  P  determine  a  plane. 

To  avoid  drawing  another  plane,  use  XY  again  to  represent 
the  plane  of  PQ  and  PR,  letting  it  cut  MN  in  the  line  CD. 

Then  since  PQ  is  ±  to  MN,  it  is  _L  to  CD.  §  430 

Therefore  PR  is  oblique  to  CD.  §  57 

Therefore  PR  is  oblique  to  MN.  §  435 

Therefore  PQ  is  the  only  ±  to  MN  at  the  point  P.  q.e.d. 

Discussion.  What  is  the  corresponding  proposition  in  plane  geometry? 


LINES  AND  PLANES  •  279 

Proposition  V.    Theorem 

437.  Through  a  given  external  point  there  can  he  drawn 
one  line  'perpendicular  to  a  given  plane,  and  only  one. 


Y 
Given  the  plane  MN  and  the  external  point  P. 

To  prove  that  there  can  he  drawn  one  line  from  P  perpen- 
dicular to  the  plane  MN^  and  only  one. 

Proof.    In  MN  draw  any  line  EH,  and  let  ZF  be  a  plane 
through  P  ±  to  EH,  cutting  MN  in  AB,  and  EH  in  C. 

Draw  PO  l.io  AB,  and  in  MN  draw  any  line  OD  from  0  to  EH. 

Produce  PO,  making  OP'  =  OP,  and  draw  PC,  PD,  P'C,  P'D. 

Since  DC  is  ±  to  XY,  APCD2indP'CD^Ye  right  angles.  §  430 

Since        the  side  DC  is  common,  and  PC  =P'C,  §  150 

.  • .  rt.  A  PC/)  is  congruent  to  rt.  A  P'CD.  §  69 

.'.PD  =  P'D.  §67 

.-.  OZ)  is  ±  to  PP' at  0.  §151 

.-.  PO  is  ±  to  MN,  being  ±  to  OD  and  AB.  §  431 

Moreover,  every  other  line  PF  from  P  to  MN  is  oblique 

to  MN.    (The  proof  is  left  for  the  student.) 

.'.PO  is  the  only  ±  from  P  to  MN.  q.e.d. 

438.  Corollary.    The  perpendicular  is   the    shortest    line 
from  a  point  to  a  plane. 

The  length  of  this  ±  is  called  the  distance  from  the  point  to  the  plane. 


280  BOOK  VI.    SOLID  GEOMETRY 

Proposition  VI.    Theorem 

439.  Oblique  lines  draivn  from  a  point  to  a  plane, 
meeting  the  plane  at  equal  distances  from  the  foot  of 
the  perpendicular,  are  equal ;  and  of  tivo  oblique  lilies, 
meeting  the  plane  at  unequal  distayices  from  the  foot 
of  the  perpendicular,  the  more  remote  is  the  greater. 


Given  the  plane  MN^  the  perpendicular  line  PO^  the  oblique  lines 
PA^  PB^  PCy  the  equal  distances  OB^  OCy  and  the  unequal  dis- 
tances OAy  OCy  with  OA  greater  than  OC. 

To  prove  that     PB  =  PCy  and  PA  >  PC, 
Proof.  In  the  A  OBP  and  OCP, 

OP  =  OP,  Iden. 

OB  =  OC,  Given 

and                                   ZB0P  =  ZP0C.  %  56 

.-.A  OBP  is  congruent  to  A  OCP.  §  69 

.'.PB  =  PC.  §67 

Let  A,  B,  and  0  lie  in  the  same  straight  line. 

Then                                  OA>OC.  Given 

.-.  OA  >  OB.  Ax.  9 

.\PA>PB.  §84 

.\PA> PC,  by  Ax.  9.  q.e.d. 

Discussion.    Compare  the  corresponding  case  in  plane  geometry. 


LINES  AND  PLANES  281 

440.  Corollary  1.  Equal  oblique  lines  drawn  from  a  point 
to  a  plane  meet  the  plane  at  equal  distances  from  the  foot  of  the 
perpendicular ;  and  of  two  unequal  oblique  lines  the  greater 
meets  the  plane  at  the  greater  distance  from  the  foot  of  the 
perpendicular. 

In  the  figure  on  page  280,  if  TB  is  given  equal  to  PC,  then  since 
TO  =  TO^  and  the  angles  at  0  are  right  angles,  what  follows  with  re- 
spect to  the  A  OBT  and  OCT'i  with  respect  to  OB  and  OG  ? 

Furthermore,  if  TA  >  PC,  how  does  FA  compare  with  PjB? 

Then  how  does  OA  compare  with  OB  ?   Why  ? 

Then  how  does  OA  compare  with  OCi 

441.  Corollary  2.  The  locus  of  a  point  equidistant  from 
all  points  on  a  circle  is  a  line  through  the  center,  perpen- 
dicular to  the  plane  of  the  circle,  ' 

In  the  figure  on  page  280,  in  order  to  prove  that  PO  is  the  required 
locus  what  must  be  proved  for -any  point  on  PO  (§  148)  ?  for  any  point 
not  on  PO  ?   Prove  both  of  these  facts. 

442.  Corollary  3.    The  locus  of  a  point  equidistant  from 

the  vertices  of  a  triangle  is  a  line  through  the  center  of  the 

circumscribed  circle,  perpendicular  to  the  plane  of  the  triangle. 

How  does  this  follow  from  Corollary  2  ? 

What  locus  is  the  line  through  the  center  of  the  inscribed  circle,  per- 
pendicular to  the  plane  of  the  triangle  ? 

443.  Corollary  4.  The  locus  of  a  point  equidistant  from 
two  given  points  is  the  plane  perpendicular  to  the  line  joiriing 
them,  at  its  mid-point. 

For  any  point  C  in  this  plane  lies  in  a  ±  to  ^Z?    ^ 
at  0,  its  mid-point  (§  430). 

Hence  how  do  CA  and  CB  compare  (§  150)  ? 

And  any  point  I)  outside  the  plane  MN  cannot  lie 
in  a  ±  to  AB  at  0.    What  may  therefore  be  said  as  to  the  distances 
from  D  to  ^  and  B  (§  150)  ? 

What  is  the  proposition  in  plane  geometry  corresponding  to  Corol- 
lary 4  ?    In  what  respect  do  the  two  proofs  differ  ? 


282 


BOOK  VI.    SOLID  GEOMETRY 


Proposition  YII.    Theorem 

444.  Two  lines  perpendicular  to  the  same  plane  are 

parallel. 

c 


B^-^^: 


N 


Given  the  lines  AB  and  CD,  perpendicular  to  the  plane  MN. 
To  prove  that     AB  and  CD  are  parallel. 

Proof.  Draw  AD  and  BD,  and  in  MN  draw  through  D 
EF  _L  to  BD,  making  DE  =  DF.    Draw  BE,  AE,  BF,  AF. 

Now  prove  that  A  BDE  and  BDF  are  congruent  (§  69),  that 
AADE  and  ADF  are  right  angles  (§  80),  and  that  BD,  CD, 
and  AD  lie  in  the  same  plane  (§  432). 

But  AB  also  lies  in  this  plane, 

and  AB  and  CD  are  both  ±  to  BD 

.'.  AB  is  II  to  CD,  by  §  95. 

445.  Corollary  1.   If  one  of  two  parallel 

dicular  to  a  plane,  the  other  is  also  perpendicular 

to  the  plane. 

For  if  through  any  point  0  of  CD  a  line  is  drawn  ±  to 
MN,  how  is  it  related  to  AB  (§  444)  ?   Now  apply  §  94. 

446.  Corollary  2.   If  two  lines  are  parallel 
to  a  third  line,  they  are  parallel  to  each  other. 

For  a  plane  MN  ±  to  CD  is  ±  to  ^5  and  EF  (§  445) 


§422 
§430 

Q.E.D. 

is  perpen- 

^        c 
o 


M 


Q 


ACE 


M 


N 


447.  Line  and  Plane  Parallel.    If  a  line  and  plane  cannot 
meet,  however  far  produced,  they  are  said  to  be  parallel. 


LINES  AND  PLANES  283 

EXERCISE  74 

1.  Why  does  folding  a  sheet  of  paper  give  a  straight  edge  ? 

2.  If  equal  oblique  lines  are  drawn  from  a  given  external 
point  to  a  plane,  they  make  equal  angles  with  lines  drawn  from 
the  points  where  the  oblique  lines  meet  the  plane  to  the  foot 
of  the  perpendicular  from  the  given  point. 

3.  If  from  the  foot  of  a  perpendicular  to  a  plane  a  line  is 
drawn  at  right  angles  to  any  line  in  the  plane,  the  line  drawn 
from  its  intersection  with  the  line  in  the  plane  to  any  point 
of  the  perpendicular  is  perpendicular  to  the  line  of  the  plane. 

4.  If  two  perpendiculars  are  drawn  from  a  point  to  a  plane 
and  to  a  line  in  that  plane  respectively,  the  line  joining  the 
feet  of  the  perpendiculars  is  perpendicular  to  the  given  line. 

5.  From  two  vertices  of  a  triangle  perpendiculars  are  let  fall 
on  the  opposite  sides.  From  the  intersection  of  these  perpen- 
diculars a  perpendicular  is  drawn  to  the  plane  of  the  triangle. 
Prove  that  a  line  drawn  to  any  vertex  of  the  triangle,  from 
any  point  on  this  perpendicular,  is  perpendicular  to  the  line 
drawn  through  that  vertex  parallel  to  the  opposite  side. 

6.  Find  the  point  in  a  plane  to  which  lines  may  be  drawn 
from  two  given  external  points  on  the  same  side  of  the  plane 
so  that  their  sum  shall  be  the  least  possible. 

Erom  one  point  A  suppose  a  ±  AO  drawn  to  the  plane  and  produced 
to  A',  making  OA'  =  OA.  Connect  A'  and  the  other  point  JB  by  a  line 
cutting  the  plane  at  P.   Then  BPA  is  the  shortest  line. 

7.  If  three  equal  oblique  lines  are  drawn  from  an  external 
point  to  a  plane,  the  perpendicular  from  the  point  to  the  plane 
meets  the  plane  at  the  center  of  the  circle  circumscribed  about 
the  triangle  having  for  its  vertices  the  feet  of  the  oblique  lines. 

8.  State  and  prove  the  propositions  of  plane  geometry  cor- 
responding to  §§  444,  445,  and  446.  Why  do  not  the  proofs 
of  those  propositions  apply  to  these  sections  ? 


284 


BOOK  VI.    SOLID  GEOMETEY 
Proposition  VIII.    Theorem 


448.  If  hvo  lines  are  parallel,  every  j^lctne  containing 
one  of  the  lines,  and  only  one,  is  p)cirallel  to  the  other  line. 


Given  the  parallel  lines  -45  and  CD^  and  the  plane  ikfiV  contain- 
ing CD  but  not  AB. 

To  prove  that    the  plane  MN  is  parallel  to  A-B. 

Proof.        AB  and  CD  are  in  the  same  plane,  A  I).  §  93 

This  plane  AD  intersects  the  plane  MN  in  CD.  Given 

Now  AB  lies  in  the  plane  AD,  however  far  produced.  §  422 

Therefore,  ii  AB  meets  the  plane  MN  at  all,  the  point  of 

meeting  must  be  in  the  line  CD.  §  422 

But  since  AB  is  II  to  CD,  Given 

.'.  AB  cannot  meet  CD.  §  93 

.'.  AB  cannot  meet  the  plane  MN. 

.-.  MN  is  II  to  AB,  by  §  447.  q.e.d. 

449.  Corollary  1.    Through  either  of  two  lines  not  in  the 

same  plane  one  plane,  and  only  one,  can  be  passed  j)Cirallel  to 

the  other. 

For  \t  AB  and  CD  are  the  Hnes,  and  we  pass  a 
plane  through  CD  and  a  hne  GE  which  is  drawn    ^j- 
parallel  to  AB,  what  can  be  said  of  the  plane  MN    / 
determined  by  CD  and  CE,  with  respect  to  the  line    Iq; 
AB?  Why  can  there  be  only  one  such  plane  ?  j^ 


LINES  AND  PLANES  285 

450.  Corollary  2.    Through  a  given  point  one  plane,  and 

only  one,  can  he  passed  parallel  to  any  two  given  lines  in  space. 

Suppose  P  the  given  point  and  AB  and  CD  the 

given  lines.    If,  now,  w^e  dravsr  through  P  the  line  A 

A'B'  parallel  to  AB,  and  the  line  CD'  parallel  to  CD,  c/f   I     p 

these  lines  w^ill  determine  the  plane  MN  (§  425).  M     \  \     \     \ 

Then  what  may  be  said  of  the  plane  MN  with  re-  /       !  |  P/     !       \ 

spect  to  the  lines  AB  and  CD  ?   Why  can  only  one    (  M' \ 

plane  be  so  passed  through  P  ?  ^ 

Discussion.  Proposition  VIII  might  of  course  be  made  more  general 
by  allowing  both  of  the  parallels  to  lie  in  the  plane  MN.  That  is,  If  two 
lines  are  parallel,  a  plane  containing  one  of  the  lines  cannot  intersect  the 
other,  although  the  other  line  might  lie  in  it. 

In  the  figure  of  Corollary  2  the  Z  D'PB'  is  sometimes  spoken  of  as  the 
angle  between  the  nonintersecting  lines  AB  and  CD,  although  this  is  not 
commonly  done  in  elementary  geometry. 

451.  Parallel  Planes.  Two  planes  which  cannot  meet,  how- 
ever far  produced,  are  said  to  be  parallel. 

EXERCISE  75 

1.  What  is  the  locus  of  a  point  in  a  plane  equidistant  from 
two  parallel  lines  ?  What  is  the  corresponding  locus  in  space, 
given  two  parallel  planes  instead  of  two  parallel  lines  ?  Draw 
the  figure,  without  proof. 

2.  Find  the  locus  in  a  plane  of  a  point  at  a  given  distance 
from  a  given  external  point.  What  is  the  corresponding  case 
of  plane  geometry  ? 

3.  If  a  given  line  is  parallel  to  a  given  plane,  the  intersection 
of  the  plane  with  any  plane  passed  through  the  given  line  is 
parallel  to  that  line. 

4.  If  a  given  line  is  parallel  to  a  given  plane,  a  line  parallel 
to  the  given  line  drawn  through  any  point  of  the  plane  lies  in 
the  plane. 


286  BOOK  VI.    SOLID  GEOMETRY 

Proposition  IX.    Theorem 

452.  Two  planes  perpendicular  to  the  same  line  are 
parallel. 


Given  the  planes  MN  and  PQ  perpendicular  to  the  line  AB. 

To  prove  that  the  planes  MN  and  PQ  are  parallel. 

Proof.    If  MN  and  PQ  are  not  parallel,  they  must  meet. 
If  they  could  meet,  we  should  have  two  planes  from  a  point 
of  their  intersection  ±  to  the  same  straight  Ime.  .       ,       «-^ 

But  this  is  impossible.  ^  \xr^Aio\^^\*  * 

.*.  MN  and  PQ  are  parallel,  by  §  451.  q.e.d.  j^  • 

EXERCISE  76  ^^      '^^^    ^""^  ^ 

1.  What  is  the  locus  of  a  point  equidistant  from  two  given 
points  A,  B,  and  also  equidistant  from  two  other  given  points 

2.  What  is  the  locus  of  a  point  at  the  distance  d  from  a 
given  plane  P,  and  at  the  distance  d'  from  a  given  plane  P'  ? 

3.  What  is  the  locus  of  a  point  at  the  distance  d  from  a 
given  plane  P,  and  equidistant  from  two  given  points  A,  B? 

4.  Find  a  point  at  the  distance  d  from  a  given  plane  P,  at 
the  distance  d'  from  a  given  plane  P',  and  equidistant  from 
two  given  points  A,  B.  Can  there  be  more  than  one  such 
point  ?    Draw  the  figure,  without  proof. 


LINES  AND  PLANES  287 

Proposition  X.    Theorem 

453.  The  intersections  of  tioo  parallel  planes  by  a  third 
plane  are  parallel  lines. 


\ 


-/^:* 


Q 
s 

Given  the  parallel  planes  MN  and  PQj  cut  by  the  plane  RS  in 
AB  and  CD  respectively. 

To  prove  that  the  intersections  AB  and  CD  are  parallel. 

Proof.  •      AB  and  CD  are  in  the  same  plane  RS.  Given 

If  AB  and  CD  meet,  the  planes  ikfiV  and  PQ  must  meet,  since 

^^  is  always  in  MN  and  CD  is  always  in  PQ.  §  422 

But  MN  and  PQ  cannot  meet.  §  451 

.\AB\^  ll,to  CD,  by  §  93.  Q.e.d. 

454.  Corollary  1.  Parallel  lines  included  between  par- 
allel planes  are  equal. 

In  the  above  figure,  suppose  AC  W  to  BD.  Then  the  plane  oi  AC  and 
BD  will  intersect  MN  and  PQ  in  lines  that  are  how  related  to  each 
other?    Then  what  kind  of  a  figure  is  ACDB? 

455.  Corollary  2.    Two  parallel  planes    are   everywhere 

equidistant  from  each  other. 

Drop  perpendiculars  from  any  points  in  MN  to  PQ.  Prove  that  these 
perpendiculars  are  parallel  and  hence  (§  454)  that  they  are  equal. 


288 


BOOK  VI.    SOLID  GEOMETRY 


Proposition  XI.    Theorem 

456.  A  line  perpendicular  to  one  of  two  jjctrallel  j^lctnes 
is  p)erpendicular  to  the  other  also. 


r 

\d 

.0 

\ 

r       ^ 

r. 

7 

F 

f"" 

\ 

/ 

— — '         ^                    ^E 

w 


Given  the  line  AB  perpendicular  to  the  plane  MN,  and  the  plane 
PQ  parallel  to  the  plane  MN. 

To  prove  that  AB  is  perpendicular  to  the  plane  PQ. 

Proof.  Pass  through  AB  two  planes  AE,  uiF,  intersecting 
MN  in  AC,  AD,  and  intersecting  PQ  in  BE,  BF,  respectively. 

Then  ^C  is  II  to  BE,  and  ^i>  is  II  to  BF.  §  453 

But  ^5  is  _L  to  ^C  and^D.  §430 

.-.  AB  is  J_  to  BE  and  BF.  .          §  97 

.'.  AB\^  1.  to  the  plane  PQ,  by  §  431.  q.e.d. 

457.  Corollary  1.  Through  a  given  point  one  plane,  and 
only  one,  can  he  passed  parallel  to  a  given  plane. 

How  is  a  plane  through  A,  A.  to  AB,  related  to  PQ  ?    Now  use  §  433. 

458.  Corollary  2.  The  locus  of  a  point  equidistant  from 
two  parallel  planes  is  a  plane  perpendicular  to  a  line  which  is 
perpendicular  to  the  planes  and  which  bisects  the  segment  cut 
off  by  them. 

459.  Corollary  3.  The  locus  of  a  point  equidistant  from 
two  parallel  lines  is  a  plane  perpendicular  to  a  line  which  is 
perpendicular  to  the  given  lines  and  which  bisects  the  segment 
cut  off'  by  them. 


LINES  AND  PLANES 


289 


Proposition  XIL    Theorem 

460.  If  two  intersecting  lines  are  each  parallel  to  a 
plane,  the  plane  of  these  lines  is  parallel  to  that  plane. 


Given  the  intersecting  lines  -4C,  AD^  each  parallel  to  the  plane 
PQ^  and  let  MiVbe  the  plane  determined  hy  AC  and  AD. 

To  prove  that        MN  is  parallel  to  PQ. 

Proof.  Draw  AB  1.  to  PQ. 

Pass  a  plane  through  AB  and  AC  intersecting  PQ  in  BE, 
and  a  plane  through  AB  and  AD  intersecting  PQ  in  BF. 

.  Then  AB\^  A- to  BE  and  BF.  §  430 

But  A  C  and  BE  lie  in  the  same  plane,  Const. 

and  A  C  cannot  meet  BE  without  meeting  the  plane  PQ,  which 

is  impossible.  §  447 

.-.BEi^  II  to  AC.  §93 

Similarly  BF  is  II  to  AD. 

.'.  ABiii  1.  to  AC  2iiidL  to  AD.  §97 

.-.  yij5  is  ±  to  the  plane  MN.  §  431 

.-.  MN  is  II  to  PQ,  by  §  452.  q.e.d. 

Discussion.  It  is  evident  that  this  proposition  does  not  depend  upon 
the  position  of  A.  For  example,  C  and  D  might  remain  where  they  are 
and  A  might  recede  a  long  distance,  AC  and  AB  becoming  more  nearly- 
parallel.  So  long  as  the  lines  intersect,  and  only  so  long,  are  we  certain 
that  the  planes  are  parallel. 


290 


BOOK  VI.    SOLID  GEOMETRY 


Proposition  XIII.    Theorem: 

^Ql.  If  two  angles  not  in  the  same  plane  have  their 
sides  respectively  parallel  and  lying  on  the  same  side  of 
the  straight  line  joining  their  vertices,  the  angles  are 
equal,  and  their  planes  are  parallel. 


T 

A 

"A,, 

p 

1                1 

/ 

A 

-"""^                                  "~" 'n' 

V 

Given  the  angles  A  and  A\  in  the  planes  MN  and  PQ  respec- 
tively, and  their  corresponding  sides  parallel  and  lying  on  the  same 
side  of  AA\ 

To  prove  that  AA  =  /-A\  and  that  MN  is  II  to  PQ. 

Proof.    Take  AD  and  A'D'  equal,  also  AC  and  A'C  equal 

Draw  DD',  CC,  CD,  CD'. 

Since  AD  is  equal  and  11  to  A'D', 

.'.  ^^'  is  equal  and  II  to  DD'. 

In  like  manner  ^^'  is  equal  and  II  to  CC. 

.'.  DD'  and  CC  are  equal, 

and  DD'  and  CC  are  parallel. 

.-.  CD=  CD'. 

.'.A  ADC  is  congruent  to  AA'D'C. 

.'.ZA  =  ZA'. 

But  MN  is  y  to  each  of  the  lines  A'C  and  A'D'.     §  448 

.-.  MN  is  II  to  PQ,  by  §  460.  Q.e.d. 

Discussion.  Why  does  not  the  proof  of  the  corresponding  proposition 
in  plane  geometry  apply  here  ? 


§130 

Ax.  8 
§44G 
§130 

§80 
§67 


LINES  AND  PLANES  291 

Proposition  XIV.    Theorem 

462.  If  two  lines  are  cut  by  three  parallel  planes,  their 
corresponding  segments  are  proportional. 


N 


7 

■'-'\       \ 

p    /  '^ 

.       \ 

/  i- 

,_\«-V  \, 

r     / 

_\\ 

/.      -:>X\ 

/     B^—         -                   ^    \ 

Given  the  lines  AB  and  CD,  cut  by  the  parallel  planes  MNy 
PQj  RSy  in  the  points  A^  E^  Bj  and  C,  F,  D,  respectively. 

To  prove  that         AE:EB=CF:  FD. 

Proof.         Draw  AD  cutting  the  plane  PQ  in  G. 

Pass  a  plane  through  AB  and  AD,  intersecting  PQ  in  the 
line  EG,  and  intersecting  RS  in  the  line  BD. 

Also  pass  a  plane  through  AD  and  CD,  intersecting  PQ  in 
the  line  GF,  and  intersecting  MN  in  the  line  A  C. 

Then  EG  is  II  to  BD, 

and  (7Fis  II  to  AC.  §453 

.•.AE:EB  =  AG:GD, 
ai^d  CF:FD  =  AG:GD.  §273 

.'.  AE:EB=CF:FD,  by  Ax.  8.  q.e.d. 

Discussion.  This  is  a  generalization  of  §  275.  It  may  be  stated  still 
more  generally,  If  two  lines  are  cut  by  any  number  of  parallel  planes,  their 
corresponding  segments  are  proportional.  In  particular,  the  case  might  be 
considered  in  which  AB  and  CD  intersect  between  the  planes. 

Why  does  not  the  proof  of  the  corresponding  case  (§  275)  in  plane 
geometry  apply  here  ? 


292  BOOK  VL    SOLID  GEOMETRY 

EXERCISE  77 

1.  Find  the  locus  of  a  line  drawn  through  a  given  point, 
parallel  to  a  given  plane. 

2.  Find  the  locus  of  a  point  in  a  given  plane  that  is  equi- 
distant from  two  given  points  not  in  the  plane. 

3.  Find  the  locus  of  a  point  equidistant  from  three  given 
points  not  in  a  straight  line. 

4.  Find  the  locus  of  a  point  equidistant  from  two  given 
parallel  planes  and  also  equidistant  from  two  given  points. 

5.  What  is  the  locus  of  a  point  in  a  plane  at  a  given  dis- 
tance from  a  given  line  in  the  plane  ?  What  is  the  locus  of 
a  point  at  a  given  distance  from  a  given  plane  ? 

6.  The  line  AB  cuts  three  parallel  planes  in  the  points  A, 
E,  B ;  and  the  line  CD  cuts  these  planes  in  the  points  C,  F,  D. 
li  AE  =  Q>  in.,  EB  =  S  in.,  and  CD  =  12  in.,  compute  CF  and  FD. 

7.  The  line  AB  cuts  three  parallel  planes  in  the  points  A, 
E,  B ;  and  the  line  CD  cuts  these  planes  in  the  points  C,  F,  D. 
lfAB  =  S  in.,  CF=5  in.,  and  CD  =  9  in.,  compute  AE  and  EB. 

8.  To  draw  a  perpendicular  to  a  given  plane  from  a  given 
point  without  the  plane. 

9."  To  erect  a  perpendicular  to  a  given  plane  at  a  given 
point  in  the  plane. 

10.  It  is  proved  in  plane  geometry  that  if  three  or  more 
parallels  intercept  equal  segments  on  one  transversal,  they 
intercept  equal  segments  on  every  transversal.  State  and  prove 
a  corresponding  proposition  in  solid  geometry. 

11.  It  is  proved  in  plane  geometry  that  the  line  joining  the 
mid-points  of  two  sides  of  a  triangle  is  parallel  to  the  third 
side.  State  and  prove  a  corresponding  proposition  in  solid 
geometry,  referring  to  a  plane  passing  through  the  mid-points 
of  two  sides  of  a  triangle. 


DIHEDRAL  ANGLES 


293 


463.  Dihedral  Angle.  The  opening  between  two  intersecting 
planes  is  called  a  dihedral  angle. 

In  this  figure  the  two  planes  AM 
and  BN  are  called  the  faces  of  the 
dihedral  angle,  and  the  line  of  inter- 
section AB  i&  called  the  edge  of  the 
angle. 

A  dihedral  angle  is  read  by  nam- 
ing the  letters  designating  its  edge, 
or  its  faces  and  edge,  or  by  a  small  letter  within.   Thus  the  dihedral 
angle  here  shown  may  be  designated  by  AB^  M-AB-N,  or  d. 

464.  Size  of  a  Dihedral  Angle.  The  size  of  a  dihedral  angle 
depends  upon  the  amount  of  turning  necessary  to  bring  one 
face  into  the  position  of  the  other. 

The  analogy  to  the  plane  angle  is  apparent,  and  is  still  further  seen 
as  we  proceed. 


465.  Adjacent  Dihedral  An- 
gles. If  two  dihedral  angles 
have  a  common  edge,  and  a 
common  face  between  them, 
they  are  said  to  be  adjacent 
dihedral  angles. 

For  example,  M-AB-N  and  N-BA-P  are  adjacent  dihedral  angles. 

466.  Right  Dihedral  Angle.  If  one  plane  meets  another  plane 
and  makes  the  adjacent  dihedral  angles  equal,  each  of  these 
angles  is  called  a  right  dihedral  angle. 

Dihedral  angles  are  said  to  be  straight,  acute,  obtuse,  reflex,  comple- 
mentary, supplementary,  conjugate,  and  vertical,  under  conditions  similar 
to  those  obtaining  with  plane  angles.  There  is  little  occasion,  however, 
to  use  any  of  these  terms  in  connection  with  dihedral  angles. 

467.  Perpendicular  Planes.  If  two  planes  intersect  and  form 
a  right  dihedral  angle,  each  of  the  planes  is  said  to  be  perpen- 
dicular to  the  other  plane. 


294  BOOK  VL    SOLID  GEOMETRY 

468.  Plane  Angle  of  a  Dihedral  Angle.    The  plane  angle  formed 
by  two  straight  lines,  one  in  each  plane,  perpen-    O 
dicular  to  the  edge  at  the  same  point,  is  called 
the  plane  angle  of  the  dihedral  angle. 

Tor  example,  Z  A  OB  is  the  plane  angle  of  the  dihedral 
angle  00'\  \i  AO  and  BO  are  each  ±  to  00'\ 

469.  Corollary.  The  plane  angle  of  a  dihedral  angle  has 
the  same  magnitude  from  whatever  point  in  the  edge  the  per- 
pendiculars are  drawn. 

How  is  O'B'  related  to  OB,  and  O'A'  to  OA  (§95)?  Then  how  is 
AA'O'B'  related  to  Z  A  OB  (§  461)  ? 

470.  Relation  of  Dihedral  Angles  to  Plane  Angles.  It  is  appar- 
ent that  the  demonstrations  of  many  properties  of  dihedral 
angles  are  identically  the  same  as  the  demonstrations  of  anal- 
ogous properties  of  plane  angles.  A  few  of  the  more  important 
propositions  will  be  proved,  but  the  following  may  be  assumed 
or  may  be  taken  as  exercises  : 

1.  If  a  plane  meets  another  plane,  it  forms  with  it  two  adjacent 
dihedral  angles  whose  sum  is  equal  to  two  right  dihedral  angles. 

2.  If  the  sum  of  two  adjacent  dihedral  angles  is  equal  to  two  right 
dihedral  angles,  their  exterior  faces  are  in  the  same  plane. 

3.  If  two  planes  intersect  each  other,  their  vertical  dihedral  angles 
are  equal. 

4.  If  a  plane  intersects  two  parallel  planes,  the  alternate-interior  dihe- 
dral angles  are  equal ;  the  exterior-interior  dihedral  angles  are  equal  ; 
and  the  two  interior  dihedral  angles  on  the  same  side  of  the  transverse 
plane  are  supplementary. 

5.  When  two  planes  are  cut  by  a  third  plane,  if  the  alternate-interior 
dihedral  angles  are  equal,  or  the  exterior-interior  dihedral  angles  are 
equal,  and  the  edges  of  the  dihedral  angles  thus  formed  are  parallel, 
the  two  planes  are  parallel. 

6.  Two  dihedral  angles  whose  faces  are  parallel  each  to  each  are 
either  equal  or  supplementary. 

7.  Two  dihedral  angles  whose  faces  are  perpendicular  each  to  each, 
and  whose  edges  are  parallel,  are  either  equal  or  supplementary. 


BIHEDKAL  ANGLES 
Proposition  XV.    Theorem 


295 


471.  Two  dihedral  angles  are  equal  if  their  plane 
angles  are  equal. 


Given  two  equal  plane  angles  ABD  and  A'B'D'  of  the  two  dihe- 
dral angles  d  and  d'. 

To  prove  that  the  dihedral  angles  d  and  d'  are  equal. 

Proof.  Apply  dihedral  angle  d'  to  dihedral  angle  d,  making 
the  plane  Z.  A'B'D'  coincide  with  its  equal  A  ABD. 

Then  since            B'C  is  ±  to  A'B'  and  D'B',  §  468 

.-.  B'C  is  _L  to  the  plane  A'B'D'.  §  431 

.-.  B'C  will  also  be  ±  to  the  plane  ABD  at  B.  Post.  5 

.-.^'C  will  fall  on  5C.  §436 

Then  the  planes  A' B'C'  and  ABC,  having  in  common  the 
two  intersecting  lines  AB  and  BC,  coincide.  §  425 

In  the  same  way  it  may  be  shown  that  the  planes  D'B'C' 
and  DBC  coincide. 

Therefore  the  two  dihedral  angles  d  and  d'  coincide  and  are 
equal.  q.e.d. 

Discussion.  May  we  have  equal  straight  dihedral  angles  ?  equal  reflex 
dihedral  angles  ?  What  is  the  authority  for  saying  that  right  dihedral 
angles  are  equal  ? 


296 


BOOK  VI.    SOLID  GEOMETRY 
Proposition  XVI.    Theorem 


472.  Two  dihedral  angles  have  the  same  ratio  as  their 
plane  angles. 


A' 


Fig.  1  Fig.  2  Fig.  3 

Given  two  dihedral  angles  BC  and.  B'C,  and  let  their  plane  angles 
be  ABD  and  A'B'D'  respectively. 

To  prove  that  ZB'C':ZBC  =  Z  A'B'D' :  Z  ABD. 

Case  1.    When  the  plane  angles  are  commensurable. 

Proof.  Suppose  the  A  ABD  and  A'B'D'  (Figs.  1  and  2)  have 
a  common  measure,  which  is  contained  m  times  in  Z.  ABD  and 
n  times  in  Z  A'B'D'. 

Then  Z  A  'B'D' :  Z  ABD  =  n:m. 

Apply  this  measure  to  /.ABD  and  Z.  A'B'D',  and  through 
the  lines  of  division  and  the  edges  BC  and  B'C  pass  planes. 

These  planes  divide  Z.BC  into  m  parts,  and  /.B'C'  into  n 
parts,  equal  each  to  each.  -   §  471 

.-./B'C  :/BC  =  n:m. 

.'.  /B'C  :  /BC  =  /A'B'D' :  /ABD,  by  Ax.  8.       Q.e.d. 

As  with  plane  angles,  there  is  also  the  case  of  incommensurables. 
Since  the  common  measure  may  be  taken  as  small  as  we  please,  it  is 
evident  that  for  practical  purposes  the  above  proof  is  sufficient.  The 
proof  for  the  incommensurable  case,  p.  297,  may  be  omitted  at  the 
discretion  of  the  teacher  without  destroying  the  sequence. 


DIHEDKAL  ANGLES  297 

Case  2.    When  the  plane  angles  are  incommensurable. 

Proof.  Divide  the  Z  ABD  into  any  number  of  equal  parts, 
and  apply  one  of  these  parts  to  the  AA'B'D'  (Figs.  1  and  3) 
as  a  unit  of  measure. 

Since  Z.ABD  and  Z.A'B'D'  are  incommensurable,  a  certain 
number  of  these  parts  will  form  the  Z.A'B'E,  leaving  a  re- 
mainder Z  EB'D',  less  than  one  of  the  parts. 

Pass  a  plane  through  B'E  and  B'C'. 

Since  the  plane  angles  of  the  dihedral  angles  A-BC-D  and 
A'-B'C'-E  are  commensurable, 

.-.  A'-B'C'-E  :  A-BC-D  =  ZA'B'E  :  Z  ABD.        Case  1 

By  increasing  the  number  of  equal  parts  into  which  /.ABD 
is  divided  we  can  diminish  the  magnitude  of  each  part,  and 
therefore  can  make  the  Z  EB'D'  less  than  any  assigned  positive 
value,  however  small. 

Hence  the  Z  EB'D'  approaches  zero  as  a  limit,  as  the  number 
of  parts  is  indefinitely  increased,  and  at  the  same  time  the  cor- 
responding dihedral  Z  £J-£'C'-Z)' approaches  zero  as  a  limit.   §  204 

Therefore  the  Z.A'B'E  approaches  the  Z.A'B'D'  as  a  limit, 
and  the  A  A'-B'C'-E  approaches  the  ZA'-B'C'-D'  as  a  limit. 

.'.  the  variable     /  ,  „^  approaches     .  ,  „^    as  a  limit, 
Z.ABD     ^^  A  ABD 

.  ,,     Z.A'-B'C'-E  ,       ZA'-B'C'-D'  ,.     .^ 

and  the  variable      ,  ,  ^^  ^   approaches      .  ,  ^^ as  a  limit. 

Z.  A-BC-D      ^^  Z.  A-BC-D 

^  ^  AA'B'E  .       ,  ,  ^     A  A'-B'C'-E  ^  ,,    , 

-^^t     /  .  r>r.    IS  always  equal  to      ,  ?   as  Z.A'B'E 

Z-ABD  Z.  A-BC-D 

varies  in  value  and  approaches  /.A'B'D'  as  a  limit.  Case  1 

AA'-B'C'-D'      ZA'B'D'    ,      ,  ^^„ 
••■    AA.BC.D    -ZaEF'''^^''"-  '•^■"- 

473.  Corollary.  The  plane  angle  of  a  dihedral  angle  may 
he  taken  as  the  measure  of  the  dihedral  angle. 


298  BOOK  VI.    SOLID  GEOMETRY 

Proposition  XVII.    Theorem 

474.  If  two  planes  are  perpendicular  to  each  other,  a 
line  draion  in  one  of  them  perpendicular  to  their  inter- 
section is  perpendicular  to  the  other. 


Given  the  planes  MN  and  PQ  perpendicular  to  each  other,  and  the 
line  CD  in  PQ  perpendicular  to  their  intersection  AB. 

To  prove  that  CD  is  perpendicular  to  the  plane  MN. 
Proof.       In  the  plane  MN  draw  DE  l.to  AB  at  D. 
Then  Z  EDC  is  a  right  angle,  §  473 

and  Z.  CDA  is  also  a  right  angle.  Given 

.-.  (7i)  is  ±  to  the  plane  MN,  by  §  431.  Q.e.d. 

475.  Corollary  1.   If  two  planes  are  perpendicular  to  each 

other^  a  perpendicular  to  one  of  them  at  any  point  of  their 

intersection  will  lie  in  the  other. 

Will  a  line  CB  drawn  in  the  plane  PQ  ±  to  ^.B  at  D  be  ±  to  the  plane 
JfiV?   How  many  Js  can  be  drawn  from  D  to  the  plane  JfiV? 

476.  Corollary  2.   If  two  planes  are  perpendicular  to  eacli 

other ^  a  perpendicular  to  the  first  from  any  point  in  the  second 

will  lie  in  the  second. 

Will  a  line  CD  drawn  in  the  plane  PQ  from  C  ±  to  AB  be  ±  to  the 
plane  MN  ?    How  many  Js  can  be  drawn  from  C  to  the  plane  MN  ? 


DIHEDRAL  ANGLES  299 

Proposition  XVIII.    Theorem 

477.  If  a  line  is  perpendicular  to  a  plane,  every  plane 
passed  through  this  line  is  perpendicular  to  the  plane. 


Given  the  line  CD  perpendicular  to  the  plane  MN  at  the  point  Z>, 
and  PQ  any  plane  passed  through  CD  intersecting  MN  in  AB. 

To  prove  that  the  plane  PQ  is  perpendicular  to  the  plane  MN. 
Proof.  Draw  DE  in  the  plane  MN±  to  AB. 

CD  is  _L  to  MN, 
.'.  CDis±toAB. 
,'.  Z.EDC  measures  Z  N-AB-P. 
But  Z  EDC  is  a  right  angle. 
.-.  PQ  is  ±  to  MN,  by  §  467. 


Since 


Given 
§430 
§473 
§430 

Q.E.D. 


EXERCISE  78 

1.  A  plane  perpendicular  to  the  edge  of  a  dihedral  angle  is 
perpendicular  to  each  of  its  faces. 

2.  If  one  line  is  perpendicular  to  another,  is  any  plane  passed 
through  the  first  line  perpendicular  to  the  second  ?    Prove  it. 

3.  If  three  lines  are  perpendicular  to  one  another  at  a  com- 
mon point,  what  is  the  relation  to  one  another  of  the  three 
planes  determined  by  the  three  pairs  of  lines  ?    Prove  it. 


300  BOOK  VI.    SOLID  GEOMETRY 

Proposition  XIX.    Theorem 

478.  If  tivo  intersecting  planes  are  each  perpendicular 
to  a  third  plane,  their  intersection  is  also  perpendicidar 
to  that  plane. 


Q 
Given  two  planes  BC  and  BD^  intersecting  in  AB^  and  each  per- 
pendicular to  the  plane  PQ. 

To  prove  that  AB  is  perpendicular  to  the  plane  PQ, 
Proof.    Let  the  plane  BC  intersect  the  plane  PQ  in  BF, 
and  let  the  plane  BD  intersect  the  plane  PQ  in  BE. 
From  any  point  A  on  AB  draw  AX  l,to  BE, 

and  from  A  draw  AY  1.  to  BF. 
Then  AX  and  A  Y  are  both  _L  to  the  plane  PQ.      §  474 
But  it  is  impossible  to  draw  two  Js  to  the  plane  PQ 

from  a  point  outside  the  plane  PQ,  §  437 

or  from  a  point  in  the  plane  PQ.  §  436 

.*.  AX  and  A  Y  must  coincide. 

But  AX  and  A  Y  can  coincide  only  if  they  lie  in  both  planes. 

And  all  points  common  to  both  planes  lie  in  AB.    §  429 

.'.  AX  and  A  Y  coincide  with  AB. 

.-.  yli3  is  _L  to  the  plane  PQ.  Q.e.d. 

Discussion.  How  does  it  appear  from  this  proof  that  AB  cannot  be 
parallel  to  PQ  ? 

The  proposition  is  illustrated  in  the  intersection  of  two  walls  of  a  room 
with  the  floor  or  the  ceiling. 


DIHEDRAL  ANGLES  301 

Proposition  XX.    Theorem 

479.  The  locus  of  a  point  equidistant  from  the  faces 
of  a  dihedral  angle  is  the  plane  bisecting  the  angle. 


Given  the  plane  AM  bisecting  the  dihedral  angle  formed  by  the 
planes  AD  and  AC. 

To  prove  that  the  plane  AM  is  the  locus  of  a  point  equi- 
distant from  the  planes  AD  and  A  C. 

Proof.    Let  EOF  be  a  plane  _L  to  ^0,  the  intersection  of  the 
planes  AD  and  AC,  at  O. 

Since  AO  is  _L  to  the  plane  EOF, 
.•.the  planes  AD,  AM,  and  ^  C  are  ±  to  the  plane  EOF.  §  477 

From  any  point  P,  in  the  intersectiofi  of  the  planes  AM  and 
EOF,  draw  PF  _L  to  OF,  and  PE  _L  to  OF. 

Then  PF  is  ±  to  AD,  and  PE  is  J_  to  A  C.  §  474 

.*.  PF  and  PE  measure  the  distances  from  the  point  P  to 
the  planes  AD  and  AC.  §  438 

Since  ^0  is  ±  to  OF,  OP,  and  OF,  §  430 

.-.  OP  bisects  ZFOE.  §473 

.*.  OP  is  the  locus  of  a  point  equidistant  from  OF  and  OF.  §  152 

.*.  AM,  which  contains  all  points  P,  is  the  locus  of  a  point 
equidistant  from  the  planes  AD  and  AC.  q.e.d. 


302 


BOOK  VI.    SOLID  GEOMETRY 


Proposition  XXI.    Theorem 

480.  Through  a  given  line  not  perpendicular  to  a  given 
plane,  one  plane  and  only  one  can  he  passed  perpen- 
dicular to  the  plane. 


Given  the  line  AB  not  perpendicular  to  the  plane  MN. 

To  prove  that  one  plane  can  he  passed  through  AB  perpen- 
dicular to  the  plane  MN^  and  only  one. 

Proof.    Erom  any  point  X  of  AB  draw  XF±  to  the  plane  MN, 
and  through  AB  and  XY  pass  a  plane  AP.       §  425 

The  plane  ^P  is  ±  to  the  plane  MN,  since  it  passes  through 
ZF,  a  line  ±  to  MN.  §  477 

Moreover,  if  two  planes  could  be  passed  through  AB  A^  to  the 
plane  MN,  their  intersection  AB  would  be  _L  to  MN.         §  478 

But  this  is  impossible,  since  ^i^  is  not  ±  to  MN.  Given 

Hence  one  plane  can  be  passed  through  AB  _L  to  the  plane 
MN,  and  only  one.  q.e.d. 

481.  Projection  of  a  Point.  The  foot  of  the  line  from  a  given 
point  perpendicular  to  a  plane  is  called  the 

projection  of  the  point  on  the  plane.  ^>^:^^  (^ 

482.  Projection  of  a  Line.    The  locus  of  the  ^ 
projections  of  the  points  of  a  line  on  a  plane    \A 
is  called  the  projection  of  the  line  on  the  plane.  ^ 


DIHEDRAL  ANGLES 


303 


Proposition  XXII.    Theorem 

483.  The  projection  of  a  straight  line  not  perpendicu- 
lar to  a  plane,  upon  that  plane,  is  a  straight  line. 


§425 

§477 
§475 


Given  the  straight  line  AB  not  perpendicular  to  the  plane  MAT, 
and  A'B'  the  projection  of  AB  upon  MN. 

To  prove  that         A'B'  is  a  straight  line. 

Proof.    From  any  point  X  of  AB  draw  .YF  _L  to  MN, 

and  pass  a  plane  AP  through  XY  and  AB. 

The  plane  ^P  is  _L  to  the  plane  MN, 

and  contains  all  the  Js  drawn  from  AB  to  MN. 

Hence  A'B'  must  be  the  intersection  of  these  two  planes. 

Therefore  A'B'  is  a  straight  line,  by  §  429.         q.e.d. 

484.  Corollary.  The  projection  of  a  straight  line  perpen- 
dicular to  a  plane,  upon  that  plane,  is  a  point. 

485.  Inclination  of  a  Line.  The  angle  which  a  line  makes 
with  its  projection  on  a  plane  is  considered  as  the  angle  which 
it  makes  with  the  plane,  and  is  called  the  inclination  of  the 
line  to  the  plane. 

Therefore  a  line  ordinarily  makes  an  acute  angle  with  a  plane,  since 
it  makes  an  acute  angle  with  its  projection  on  the  plane.  The  cases  of 
perpendicular  and  parallel  lines  have  already  been  considered. 


304  BOOK  VI.    SOLID  GEOMETRY 

Proposition  XXIII.    Theorem 

486.  The  acute  angle  which  a  line  makes  with  its 
projection  upon  a  plane  is  the  least  angle  which  it 
makes  with  any  line  of  the  plane. 


N 

Given  the  line  AB  meeting  the  plane  MN at  A,  AB'  being  the  pro- 
jection of  AB  upon  the  plane  ikfiST,  and  AD  being  any  other  line  drawn 
through  A  in  the  plane  MN. 

To  prove  that  Z  B'AB  is  less  than  Z  DAB. 

Proof.    Make  AD  equal  to  AB',  and  draw  BB'  and  BD. 
Then  in  A  BAB'  and  BAD, 

AB  =  AB,  Iden. 

AB'  =  AD,  Const. 

and  BB'<BD.  §438 

.-.  Z  B'AB  <  Z  DAB,  by  §  116.  Q.e.'d. 

Discussion.  Since  Z  B'AB  is  the  least  angle  that  AB  makes  with  any 
line  of  the  plane,  how  does  ZBAC  compare  with  the  angles  that  AB 
makes  with  other  lines  of  the  plane  ?  State  the  general  proposition 
involved  in  the  answer. 

If  AB  is  parallel  to  the  plane,  what  interpretation  may  be  given  to 
the  proposition  ? 

If  AB  is  perpendicular  to  the  plane,  what  interpretation  may  be  given 
to  the  proposition  ? 

As  AD  swings  around  from  the  position  AB'  to  the  position  J. C,  what 
kind  of  change  takes  place  in  the  angle  DAB  ? 


DIHEDRAL  ANGLES  305 

EXERCISE  79 

1.  Describe  the  position  of  a  segment  of  a  line  relative  to  a 
given  plane  if  the  projection  of  the  segment  on  the  plane  is 
equal  to  its  own  length. 

2.  From  a  point  A,  4:  in.  from  a  plane  MN,  an  oblique  line 
AC  5  in.  long  is  drawn  to  the  plane  and  made  to  turn  around 
the  perpendicular  AB  dropped  from  A  to  the  plane.  Find  the 
area  of  the  circle  described  by  the  point  C. 

3.  From  a  point  ^j  8  in.  from  a  plane  MN,  a  perpendicular  AB 
is  drawn  to  the  plane ;  with  5  as  a  center  and  a  radius  equal 
to  6  in.,  a  circle  is  described  in  the  plane ;  at  any  point  C  on 
this  circle  a  tangent  CD  is  drawn  24  in.  in  length.  Find  the 
distance  from  A  to  D. 

4.  Equal  lines  drawn  from  a  given  external  point  to  a  given 
plane  are  equally  inclined  to  the  plane. 

5.  If  three  equal  lines  are  drawn  to  a  plane  from  an  exter- 
nal point,  the  perpendicular  from  the  point  to  the  plane  deter- 
mines the  center  of  the  circle  circumscribed  about  the  triangle 
determined  by  the  planes  of  the  three  lines. 

6.  Three  lines  not  in  the  same  plane  meet  in  a  point.  How 
shall  a  line  be  drawn  so  as  to  make  equal  angles  with  all  three 
of  these  lines  ? 

7.  From  a  point  P  two  perpendiculars  PX  and  PY  are  drawn 
to  two  planes  MTV  and  ^ C  which  intersect  in  AB.  From  Y  a 
perpendicular  YZ  is  drawn  to  MN.  Prove  that  the  line  XZ  is 
perpendicular  to  ^i^. 

8.  If  the  length  of  the  shadow  of  a  tree  standing  on  level 
ground  exceeds  the  height  of  the  tree,  the  angle  made  by  the 
sun  above  the  horizon  must  be  less  than  what  known  angle  ? 

9.  Find  the  locus  of  a  point  at  a  given  distance  from  a  given 
plane  and  equidistant  from  two  given  points  not  in  the  plane. 


306  BOOK  VI.    SOLID  GEOMETRY 

Proposition  XXIV.    Theorem 

487.  Between  two  lines  not  in  the  same  plane  there 
can  he  one  common  perpendicular,  and  only  one. 


Given  AB  and  CD,  two  lines  not  in  the  same  plane. 

To  prove  that  there  can  he  one  common  perpendicular^  and 
one^  between  AB  and  CD. 

Proof.    Through  any  point  A  of  AB  draw  AG  W  to  DC. 

Let  MN  be  the  plane  determined  hj  AB  and  AG.   %  425 
Then  the  plane  MN  is  II  to  DC.  §  448 

Through  DC  pass  the  plane  PQ  ±  to  the  plane  MN.  §  480 
Then  DC  cannot  meet  D'C,  since  it  is  II  to  the  plane  MN  and 
lies  in  the  plane  PQ.  §  422 

.-.  i)C  is  II  to  D^C\  §  93 

.-.  if  AB  is  II  to  D'C  it  must  be  II  to  DC.  §  446 

But  AB  is  not  II  to  DC,  for  they  are  not  in  the  same  plane.   Given 
.'.  AB  must  intersect  D'C"  at  some  point  as  C. 
Draw  C'C  ±  to  the  plane  MN. 
Then  C'C  is  ±  to  AB  and  to  i>'C'.  §  430 

Since        C'C  is  ±  to  D'C,  and  lies  in  plane  PQ,  §  475 

.-.  C'C  is  A.  to  DC.  §  97 

Therefore  one  common  perpendicular  can  be  drawn. 
It  remains  to  be  proved  that  no  other  can  be  drawn. 


DIHEDEAL  ANGLES  307 

If  it  were  possible  that  another  common  perpendicular  could 
be  drawn,  we  might  suppose  EA  to  be  J_  to  both  AB  and  CD. 

Then  EA  would  be  _L  to  AG,  §  97 

and  therefore  EA  would  be  ±  to  the  plane  MN.  §  431 

Draw^^'±toZ>'C'. 

Then  EE'  is  _L  to  the  plane  MN.  §  474 

But  this  is  impossible,  if  EA  is  also  ±  to  the  plane  MN.  §  437 

Hence  the  supposition  that  there  is  a  second  common  per- 
pendicular, EA,  leads  to  an  absurdity. 

Therefore  there  can  be  one  common  perpendicular,  and  only 
one,  between  AB  and  CD.  q.e.d. 

488.  Corollary.  The  common  perpendicular  between  two 
lines  not  in  the  same  plane  is  the  shortest  line  joining  them. 

How  does  CC  compare  in  length  with  EE'  ?   Why  ? 
How  does  EE'  compare  in  length  with  EA  ? 

EXERCISE  80 

1.  Parallel  lines  have  parallel  projections  on  a  plane. 

2.  If  two  planes  are  perpendicular  to  each  other,  any  line 
perpendicular  to  one  of  them  is  how  related  to  the  other  ? 

3.  If  three  lines  passing  through  a  given  point  P  are  cut  by 
a  fourth  line  that  does  not  pass  through  P,  the  four  lines  all 
lie  in  the  same  plane. 

4.  Seven  lines,  no  three  of  which  lie  in  the  same  plane, 
pass  through  the  same  point.  How  many  planes  are  deter- 
mined by  these  lines  ? 

5.  A  cubical  tank  10  in.  deep  contains  water  to  a  depth  of 
7  in.  A  foot  rule  is  placed  obliquely  on  the  bottom  so  as  just 
to  reach  the  top  edge  of  the  tank.  Make  a  sketch  of  the  tank, 
and  compute  the  length  of  the  rule  covered  by  water. 


308  BOOK  VI.    SOLID  GEOMETRY 

489.  Polyhedral  Angle.  The  opening  of  three  or  more  planes 
which  meet  at  a  common  point  is  called  a  ^jolyhedral  angle. 

The  common  point  F  is  called  the  xieriex  of  the  angle  ;  y 

the  intersections  F-4,  YB.,  etc.,  of  the  planes  are  called              /y\ 
the  edges;  the  portions  of  the  planes  lying  between  the          /Y    \ 
edges  are  called  the  faces;   and  the  angles  formed  by    /f-yC       \ 
adjacent  edges  are  called  the  face  angles.  /- --Ao 

Every  two  adjacent  edges  form  a  face  angle,  and  every 
two  adjacent  faces  form  a  dihedral  angle.   The  face  angles  and  dihedral 
angles  are  the  parts  of  the  polyhedral  angle. 

490.  Size  of  a  Polyhedral  Angle.  The  size  of  a  polyhedral 
angle  depends  upon  the  relative  position  of  its  faces,  and  not 
upon  their  extent. 

491.  Convex  and  Concave  Polyhedral  Angles.  A  polyhedral 
angle  is  said  to  be  convex  or  concave  according  as  a  section 
made  by  a  plane  that  cuts  all  its  edges  at  other  points  than 
the  vertex  is  a  convex  or  concave  polygon. 

Only  convex  polyhedral  angles  are  considered  in  this  work. 

492.  Classes  of  Polyhedral  Angles.  A  polyhedral  angle  is  called 
a  trlliedral  angle  if  it  has  three  faces,  a  tetrahedral  angle  if  it 
has  four  faces,  and  so  on. 

Other  names,  like  pentahedral,  hexahedral,  heptahedral,  etc.,  for 
angles  with  5,  6,  7,  etc.,  faces,  are  rarely  used. 

A  polyhedral  angle  is  designated  by  a  letter  at  the  vertex,  or  by  let- 
ters representing  the  vertex  and  all  the  faces  taken  in  order.  Thus,  in 
the  above  figure  the  trihedral  angle  is  designated  by  V  or  by  V-ABC. 
A  tetrahedral  angle  would  be  designated  by  V  or  by  V-ABC D. 

493.  Equal  Polyhedral  Angles.    If  >F               jV' 

the  corresponding  parts  of  two  poly-  yv  \              //  \ 

hedral  angles  are  equal  and  are  ar-  a/C/      \     ^aV       \ 

ranged  in  the  same  order,  the  poly-  y_lll^-J^     \^_"J1--_J^/ 

hedral  angles  are  said  to  be  equal.  ' 

Thus  the  angles  V-ABC  and  V'-A'B'C  are  equal.  Equal  polyhedral 
angles  may  evidently  be  made  to  coincide  by  superposition. 


POLYHEDRAL  ANGLES  309 

Proposition  XXV.    Theorem 

494.  Tfie  sum  of  any  two  face  angles  of  a  trihedral 
angle  is  greater  than  the  third  face  angle. 


'r 

Given  the  trihedral  angle  V-XYZ^  with  the  face  angle  XFZ  greater 
than  either  of  the  face  angles  XVY  or  YVZ. 

To  prove  that  Z  XVY-\-  Z  YVZ  is  greater  tliayi  Z  XVZ. 

Proof.    In  the  Z  XVZ  draw  VW,  making  AXVW  =  Z.  XVY. 
Through  any  point  D  of  VW  draw  ADC  in  the  plane  XVZ. 
On  FF  take  VB  equal  to  VD. 
Pass  a  plane  through  the  line  AC  and  the  point  B. 
Then  since  ^F  =  ^F,  VD  =  VB,  and  ZAVD  =  ZAVB, 

.-.AAVDis  congruent  to  A  ^  VB.  §  68 

.'.AD  =  AB.  §67 

In  the  A ^5C,  AB^BOAC.  §112 

Since  AB  =  AD,      .'.  BC  >DC.  Ax.  6 

In  the  ABVC  and  DVC, 

VC  =  VC,  and  VB  =  VD,  but  BC  >  DC 

.'.  ZBVCis  greater  than  Zi>FC.  §  116 

.-.  Z  A  VB  -\-ZBVC  is  greater  than  ZAVD-\-  ZDVC.  Ax.  6 

But    '  ZAVD  +  ZDVC  =  ZAVC.  Ax.  11 

.'.  ZAVB-^ZBVC  is  greater  than  Z  A  VC.       Ax.  9 

That  is,  ZXVY-^Z  YVZ  is  greater  than  Z  X  VZ.        q.  e.  d. 


310  BOOK  VI.    SOLID  GEOMETRY 

Proposition  XXVI.    Theorem 

495.  The  sum  of  the  face  angles  of  any  convex  pohj- 
hedral  angle  is  less  than  four  right  angles. 

V 


Given  a  convex  polyhedral  angle  F,  all  of  its  edges  being  cut  by 
a  plane  making  the  section  ABCDE. 

To  prove  that  Z  A  VB  -{-  Z.B  VC,  etc,  is  less  than  four  rt.  A. 

Proof.  From  any  point  P  within  the  polygon  draw  PA,  PB, 
PC,  PD,  PE. 

The  number  of  the  A  having  the  common  vertex  P  is  the 
same  as  the  number  having  the  common  vertex  V. 

Therefore  the  sum  of  the  A  of  all  the  A  having  the  common 
vertex  V  is  equal  to  the  sum  of  the  A  of  all  the  A  having  the 
common  vertex  P. 

But  in  the  trihedral  A  formed  Sit  A,  B,  C,  etc., 

Z  EAV-h  A  BAV  is  greater  than  ABAE, 

A  VBA  A- AC  BV  is  greater  than  A  CBA ,  etc.       §  494 

Hence  the  sum  of  the  A  at  the  bases  of  the  A  whose  com- 
mon vertex  is  V  is  greater  than  the  sum  of  the  A  at  the  bases 
of  the  A  whose  common  vertex  is  P.  Ax.  7 

Therefore  the  sum  of  the  A  at  the  vertex  V  is  less  than  the 
sum  of  the  A  at  the  vertex  P.  Ax.  7 

But  the  sum  of  the  Z  at  P  is  equal  to  4  rt.  A.  §  41 

Therefore  the  sum  of  the  zi  at  F  is  less  than  4  rt.  A.    Q.  e.  d. 


POLYHEDEAL  ANGLES 


311 


496.  Symmetric  Polyhedral  Angles.  If  the  faces  of  a  poly- 
hedral angle  V-ABCD  are  produced  through  the  vertex  V, 
another  polyhedral  angle  V-A'B'C'D'  is  formed,  symmetric  with 
respect  to  Z  V-ABCD. 

The  face  angles  AVE,  BVC, 
etc.,  are  equal  respectively  to 
the  face  angles  A'VB',  B'VC\ 
etc.  (§  60). 

Also  the  dihedral  angles  VA, 
F-B,  etc.,  are  equal  respectively 
to  the  dihedral  angles  VA\  VB\ 
etc.  (§470).  (The  second  figure 
shows  a  pair  of  these  vertical 
dihedral  angles.) 

Looked  at  from  the  point  F,  the  edges  of  Z  V-ABCD  are  arranged 
from  left  to  right  (counterclockwise)  in  the  order  VA,  FJ5,  FC,  FD,  but 
the  edges  of  Z  V-A'B'C'D'  are  arranged  from  right  to  left  (clockwise) 
in  the  order  VA'^  VB'^  VC,  VD';  that  is,  in  an  order  the  reverse  of  the 
order  of  the  edges  in  Z  V-ABCD.    Therefore, 

Two  symmetric  polyhedral  angles  have  all  their  parts  equal  each  to  each 
hut  arranged  in  reverse  order. 

497.  Symmetric  Polyhedral  Angles  not  Superposable.  In  gen- 
eral, two  symmetric  polyhedral  angles  are  not  superposable. 
Thus,  if  the  trihedral  angle  V-A'B'C  is  made  to 
turn  180°  about  XY,  the  bisector  of  the  angle 
CVA',  then  VA'  will  coincide  with  VC,  VC  with 
VA,  and  the  face  A'VC  with  A  VC ;  but  the  di- 
hedral angle  F^4,  and  hence  the  dihedral  angle 
VA',  not  being  equal  to  VC,  the  plane  A'VB'  will 
not  coincide  with  B  VC ;  and,  for  a  similar  reason, 
the  plane  C'VB'  will  not  coincide  with  AVB.  Hence  the  edge 
VB'  takes  some  position  VB"  not  coincident  with  VB ;  that  is, 
the  trihedral  angles  are  not  superposable. 

An  analogous  case  is  seen  in  a  pair  of  gloves.  All  the  parts  of  one 
are  equal  to  the  corresponding  parts  of  the  other,  but  the  right-hand 
glove  will  not  fit  the  left  hand. 


312  BOOK  VL    SOLID  GEOMETRY 

Proposition  XXVII.    Theorem 

498.  TiDO  trihedral  angles  are  equal  or  symmetric 
when  the  three  face  angles  of  the  one  are  equal  respec- 
tively to  the  three  face  angles  of  the  other. 


B'  E'      ■  a!        B  E  A     A'  E'B' 

Given  the  trihedral  angles  V  and  V\  the  angles  BVA,  CVA,  CVB 
being  equal  respectively  to  the  angles  B'V'A',  C'V'A'y  C'V^B'. 

To  prove  that  the  angles  V  and  V  are  equal  or  symmetric. 

Proof.    On  the  edges  of  these  angles  take  the  six  equal  seg- 
ments VA,  VB,  VC,  V'A',  V'B',  V'C. 

Draw  AB,  BC,  CA,  A'B',  B'C,  C'A'. 

The  isosceles  A  BA  V,  CA  V,  CB  V  are  congruent  respectively 
to  the  isosceles  AB'A'V',  C'A'V,  C'B'V.  §  68 

.-.  AB,  BC,  CA  are  equal  respectively  to  A'B',  B'C,  C'A'.  §  67 

.'.ABAC  is  congruent  to  A  B'A'C.  §  80 

From  any  point  D  in  VA  draw  J)E  in  the  face  A  VB  and  DF 
in  the  face  A  VC,  each  ±.  to  VA. 

These  lines  meet  AB  and  AC  respectively. 

{For  the  AVAB  and  VAC  are  acute,  each  being  one  of  the  equal 
A  of  an  isosceles  A.) 

Draw  EF. 
On  A'V  take  A'J)'  equal  to  AB. 


POLYHEDRAL  ANGLES  313 

Draw  7)'^'  in  the  face  A'V'B'  and  D'F'  in  the  face  A'V'C,  each 
±  to  V'A',  and  draw  E'F'. 

Then  since                      AD  =  A'D',  Const. 

and                                 ZDAE  =  ZD'A'E',  §67 

.'.Yt  A  ADE  is  congruent  to  rt.  AA'D'E'.  §  72 

.-.  ^£;  =  ^'^',  SLud  DE  =  D'E'.  §  67 

In  like  manner    AF=A'F',  and  Z)F=  X>'F'. 

Furthermore,  since  it  has  been  proved  that 

A^^C  is  congruent  to  AB'A'C, 

.\ZCAB  =  ZC'A'B'.  §67 

.'.  AAFE  is  congruent  to  A  A  'F'E'.  §  68 

.•.EF=E'F'.  §67 

.-.A  EDF  is  congruent  to  A  E'D'F'.  §  80 

.-.  Z  FDE  =  Z  F'D'E'.  §  67 

.-.  dihedral  ZVA=  dihedral  Z  V'A'.  .  §  473 

(For  A  FDE  and  F'D'E\  the  measures  of  these  dihedral  A,  are  equal.) 

In  like  manner  it  may  be  proved  that  the  dihedral  angles 
VB  and  VC  are  equal  respectively  to  the  dihedral  angles  V'B' 
and  V'C. 

.'.  the  trihedral  angles  V  and  V'  are  equal,         §  493 

or  else  they  are  symmetric,  by  §  496.  q.e.d. 

This  demonstration  applies  to  either  of  the  two  figures  denoted  by 
V'-A'B'C\  which  are  symmetric  with  respect  to  each  other.  If  the  first 
of  these  figures  is  taken,  V  and  V  are  equal.  If  the  second  is  taken, 
V  and  V  are  symmetric. 

499.  Corollary.    If  two  trihedral  angles  have  the  three 

face  angles   of  the  one  equal  respectively  to  the  three  face 

angles  of  the  other ^  then  the  dihedral  angles  of  the  one  are 

equal  respectively  to  the  dihedral  angles  of  the  other. 

For  whether  the  trihedral  angles  are  equal  or  symmetric,  as  stated  in 
the  proposition,  the  dihedral  angles  are  equal  (§§  493,  496). 


314  BOOK  VI.    SOLID  GEOMETKY 

EXERCISE  81 

1.  Find  the  locus  of  a  point  in  a  space  of  three  dimensions 
equidistant  from  two  given  intersecting  lines. 

2.  Find  a  point  at  equal  distances  from  four  points  not  all 
in  the  same  plane. 

3.  Two  dihedral  angles  which  have  their  edges  parallel  and 
their  faces  perpendicular  are  equal  or  supplementary. 

4.  The  projections  on  a  plane  of  equal  and  parallel  line- 
segments  are  equal  and  parallel. 

5.  Two  trihedral  angles  are  equal  when  two  dihedral  angles 
and  the  included  face  angle  of  the  one  are  equal  respectively 
to  two  dihedral  angles  and  the  included  face  angle  of  the  other, 
and  are  similarly  placed. 

6.  Two  trihedral  angles  are  equal  when  two  face  angles  and 
the  included  dihedral  angle  of  the  one  are  equal  respectively 
to  two  face  angles  and  the  included  dihedral  angle  of  the  other, 
and  are  similarly  placed. 

7.  If  the  face  angle  A  VB  of  the  trihedral  angle  V-ABC  is 
bisected  by  the  line  VD,  the  angle  CVD  is  less  than,  equal  to, 
or  greater  than  half  the  sum  of  the  angles  A  VC  and  B  VC, 
according  as  Z  C  VD  is  less  than,  equal  to,  or  greater  than  90°. 

8.  If  two  face  angles  of  a  trihedral  angle  are  equal,  the 
dihedral  angles  opposite  them  are  equal. 

9.  A  trihedral  angle  having  two  of  its  face  angles  equal  is 
superposable  on  its  symmetric  trihedral  angle. 

10.  Find  the  locus  of  a  point  equidistant  from  the  three  edges 
of  a  trihedral  angle. 

11.  Find  the  locus  of  a  point  equidistant  from  the  three  faces 
of  a  trihedral  angle. 

12.  The  planes  that  bisect  the  dihedral  angles  of  a  trihedral 
angle  meet  in  a  straight  line. 


EXEECISES  315 

EXERCISE  82 

Problems  of  Computation 

1.  From  a  point  P,  4  in.  from  a  plane,  a  line  PX  is  drawn 
meeting  the  plane  at  X.  If  PX  is  5  in.,  what  is  the  length  of 
the  locus  of  X  in  the  plane  ? 

2.  From  a  point  P,  5  in.  from  a  plane,  a  line  PX  is  drawn 
meeting  the  plane  at  X.  If  PX  is  12  in.,  what  area  is  inclosed 
in  the  plane  by  the  locus  oiX?  Answer  to  two  decimal  places. 

3.  The  base  AB  of  the  isosceles  triangle  ABC  in  the-  plane 
MN  is  6  in.,  and  the  perimeter  of  the  triangle  is  20  in.  If  the 
triangle  revolves  about  its  base  as  an  axis,  what  is  the  greatest 
distance  from  the  plane  that  is  reached  by  C  ?  Answer  to  three 
decimal  places. 

4.  Two  points  A  and  B  are  4  in.  apart.  A  point  P  moves  so 
as  to  be  constantly  5  in.  from  each  of  these  points.  Find  the 
length  of  the  locus  of  P.    Answer  to  three  decimal  places. 

5.  Two  parallel  planes  MN  and  PQ  are  cut  by  a  third  plane 
RS  so  as  to  make  one  of  the  dihedral  angles  27°  15'  30".  Find 
the  other  dihedral  angles. 

6.  Two  lines  are  cut  by  three  parallel  planes.  The  segments 
cut  from  one  line  are  3  in.  and  5^  in.,  and  those  cut  from  the 
other  line  are  7|  in.  and  x.    Find  the  value  of  x. 

7.  Two  given  planes  are  at  right  angles  to  each  other.  A 
point  Z  is  8  in.  from  each  plane.  How  far  is  X  from  the  edge 
of  the  right  dihedral  angle  ? 

8.  What  is  the  length  of  the  projection  on  a  plane  of  a  line 
whose  length  is  10  V2,  the  inclination  of  the  line  to  the  plane 
being  45°  ? 

9.  From  the  external  point  P  a  perpendicular  PP',  9  in.  long, 
is  drawn  to  a  plane  MN.  From  P  the  line  PQ  is  drawn  to  the 
plane  making  the  angle  P'PQ  equal  to  30°.  Find  the  length  of 
the  projection  of  PQ  on  the  plane  MN. 


316  BOOK  VL    SOLID  GEOMETRY 

EXERCISE  83 

Review  Questions 

1.  How  many  and  what  conditions  determine  a  straight  line  ? 
How  many  and  what  conditions  determine  a  plane  ? 

2.  What  simple  numerical  test,  following  the  measurement 
of  certain  lengths,  determines  whether  or  not  one  line  is  perpen- 
dicular to  another  ?  a  line  is  perpendicular  to  a  plane  ? 

3.  How  many  planes  can  be  passed  through  a  given  line 
perpendicular  to  a  given  plane  ?  Is  this  true  for  all  positions 
of  the  given  line  ? 

4.  Through  a  given  point  how  many  lines  can  be  drawn 
parallel  to  a  given  line  ?  parallel  to  a  given  plane  ?  Through 
a  given  point  how  many  planes  can  be  passed  parallel  to  a 
given  line  ?  parallel  to  a  given  plane  ? 

5.  What  is  the  locus,  in  a  line,  of  a  point  equidistant  from 
two  given  points  ?  in  a  plane  ?  in  a  space  of  three  dimensions  ? 

6.  What  is  the  locus,  in  a  plane,  of  a  point  equidistant  from 
two  intersecting  lines  ?  State  a  corresponding  proposition  for 
solid  geometry. 

7.  What  may  be  said  of  two  lines  in  one  plane  perpendicular 
to  the  same  line  ?  State  two  corresponding  propositions  for 
solid  geometry.  Does  one  of  these  propositions  state  that  two 
planes  perpendicular  to  the  same  plane  are  parallel  ? 

8.  What  may  be  said  of  a  line  perpendicular  to  one  of  two 
parallel  lines  ?  State  two  corresponding  propositions  for  solid 
geometry.  Is  a  plane  perpendicular  to  one  of  two  parallel 
planes  perpendicular  to  the  other  ? 

9.  If  a  line  is  perpendicular  to  a  plane,  what  may  be  said 
of  every  plane  passed  through  this  line  ?  Does  a  true  prop- 
osition result  from  changing  the  word  "perpendicular"  to 
"parallel"  in  this  statement? 


BOOK  VII 

POLYHEDRONS,  CYLINDERS,  AND  CONES 

500.  Polyhedron.  A  solid  bounded  by  planes  is  called  a  ][)oly- 
hedron. 

For  example,  the  figures  on  pages  317  and  318  are  polyhedrdns. 

The  bounding  planes  are  called  the  faces  of  the  polyhedron,  the  in- 
tersections of  the  faces  are  called  the  edges  of  the  polyhedron,  and  the 
intersections  of  the  edges  are  called  the  vertices  of  the  polyhedron. 

A  line  joining  any  two  vertices  not  in  the  same  face  is  called  a 
diagonal  of  the  polyhedron. 

The  plural  of  polyhedron  is  polyhedrons  or  poly hedra. 

501.  Section  of  a  Polyhedron.  If  a  plane  passes  through  a 
polyhedron,  the  intersection  of  the  plane  with  such  faces  as  it 
cuts  is  called  a  section  of  the  polyhedron. 

502.  Convex  Polyhedron.  If  every  section  of  a  polyhedron 
is  a  convex  polygon,  the  polyhedron  is  said  to  be  convex. 

Only  convex  polyhedrons  are  considered  in  this  v^ork. 

503.  Prism.    A  polyhedron   of  which  two  faces   are   poly- 
gons in  parallel  planes,  and  the  other  faces 
are  parallelograms,  is  called  a.  prism. 

The  parallel  polygons  are  called  the  bases  of  the 
prism,  the  parallelograms  are  called  the  lateral 
faces^  and  the  intersections  of  the  lateral  faces 
are  called  the  lateral  edges. 

The  sum  of  the  areas  of  the  lateral  faces  is 
called  the  latei^al  area  of  the  prism. 

The  lateral  edges  of  a  prism  are  equal  (§  125). 

504.  Altitude  of  a  Prism.  The  perpendicular  distance  be- 
tween the  planes  of  the  bases  of  a  prism  is  called  its  altitude. 

317 


318 


BOOK  VII.    SOLID  GEOMETRY 


Uiulit  I'ri,>^i 


505.  Right  Prism.    A  prism  whose  lateral  edges  are  per- 
pendicular to  its   bases   is  called  a  right 
prism. 

The  lateral  edges  of  a  right  prism  are  equal  to 
the  altitude  (§455). 

506.  Oblique  Prism.  A  prism  whose  lat- 
eral edges  are  oblique  to  its  bases  is  called 
an  oblique  prism. 

507.  Prisms  classified  as  to  Bases.  Prisms 
are  said  to  be  triangular,  quadrangular, 
and  so  on,  according  as  their  bases  are 
triangles,  quadrilaterals,  and  so  on. 

508.  Right  Section.  A  section  of  a  prism 
made  by  a  plane  cutting  all  the  lateral  edges 
and  perpendicular  to  them  is  called  a  right 

section.  Oblique  Triangular  Prism 

In  the  case  of  oblique  prisms  it  is  sometimes  necessary  to  produce 
some  of  the  edges  in  order  that  the  cutting  plane  may  intersect  them. 


Right  Section  uf  a  I'risin  Truncated  Prism 

509.  Truncated  Prism.  The  part  of  a  prism  included  between 
the  base  and  a  section  made  by  a  plane  oblique  to  the  base  is 
called  a  truncated  j^risTn. 


PRISMS  319 

Proposition  I.    Theorem 

510.  The  sections  of  a  prism  made  hy  parallel  planes 
cutting  all  the  lateral  edges  are  congruent  polygons. 


Given  the  prism  PR  and  the  parallel  sections  AD^  A^D'  cutting 
all  the  lateral  edges. 

To  prove  that     AD  is  congruent  to  A'D'. 

Proof.    AB  is  II  to  A'B',  EC  is  II  to  B'C\  CD  is  II  to  C'D\ 

and  so  on  for  all  the  corresponding  sides.  §  453 

.'.  AB  =  A'B',  BC  =  B'C',  CD  =  C'D', 
and  so  on  for  all  Ihe  corresponding  sides,  §  127 

and  Z  CBA  =  Z  C'B'A  \Z.DCB  =  Z.  D'C'B', 

and  so  on  for  all  the  corresponding  angles.         §  461 
.*.  AD  is  congruent  to  A'D',  by  §  142.  Q-e.d. 

Discussion.  Is  the  proof  the  same  whether  or  not  the  two  parallel 
planes  are  parallel  to  the  hases  ? 

If  the  sections  are  all  parallel  to  the  bases,  are  they  also  congruent  to 
the  bases  ? 

Would  the  proposition  be  true  if  the  prism  were  concave  instead  of 
convex  ? 

Suppose  the  bases  were  squares,  what  would  be  known  as  to  the  form 
of  the  sections  ? 

511.  Corollary.  Every  section  of  a  prism  made  hy  a 
plane  parallel  to  the  base  is  congruent  to  the  base ;  and  all 
right  sections  of  a  prism  are  congruent. 


320  BOOK  VII.    SOLID  GEOMETRY 

Proposition  II.    Theorem 

512.  The  lateral  area  of  a  prism  is  equal  to  the 
product  of  a  lateral  edge  by  the  perimeter  of  a  right 
section. 


Given  VWXYZ  a  right  section  of  the  prism  AD\  I  the  lateral 
area,  e  a  lateral  edge,  and  p  the  perimeter  of  the  right  section. 

To  prove  that  I  =  ep. 

Proof.  AA'  =  BB'=CC'  =  DD'  =  EE'=:e.  §503 

Furthermore,  VW  is  _L  to  BB\  WX  to  CC\  XY  to  DD\  YZ 

to  EE\  and  ZVto  AA'.  §  508 

.-.  the  area  of  O  AB'  =  BB'  xVW=eX  VW,  §  322 

the  area  of  O  BC'  =CC'x  WX  =  ex  WX, 

the  area  of  O CD'  =  DD' X  XY  ==  e  X  XY,  and  so  on. 

But  I  is  equal  to  the  sum  of  these  parallelograms.  §  503 

.-.  I  =  e  (VW  +  WX  -j-  XY -{-  YZ  -\-  ZV).  Ax.  1 

But  VW-{-WX-{-XY-\-YZ-}-ZV=p.  Ax.  11 

.'.  1  =  ep,hj  Ax.  9.  Q.E.D. 

513.  Corollary.    The   lateral   area   of  a   right  prism   is 

equal  to  the  product  of  the  altitude  hy  the  perimeter  of  the  base. 

Tor  how  would  p  then  compare  with  AB  -\-  BC  -^  CD  -\-  BE  +  EA? 
The  truth  of  the  corollary  is  easily  seen  by  imagining  the  right  prism 
laid  on  one  of  its  lateral  faces,  and  the  surface  as  it  were  unrolled. 


PRISMS  321 

EXERCISE  84 

Find  the  lateral  areas  of  the  right  prisms  whose  altitudes 
and  perimeters  of  bases  are  as  follows : 

1.  6^  =  18  in.,  p  =  29  in.        4.  a  ==  1  ft.  7  in.,  p  =  2  ft.  9  in. 

2.  a  =  22  in.,  p  =  37  in.        5.  r^  =  3  ft.  8  m.,p  =  5  ft.  7  in. 

3.  a  =  4.25  in.,^  =3  6.75  in.   6.  a  =12  ft.  2  in., 7;  ==  27  ft.  9  in. 

Find  the  lateral  areas  of  the  prisms  whose  lateral  edges  and 
perimeters  of  right  sections  are  as  follows : 

7.  e  =  17  in.,  p  =  27  in.       10.  e  ==  1  ft.  3  in.,  ^  =  2  ft.  3  in. 

8.  e  =23  in., p  =  35  in.       11.  e  =  2  ft.  7  in,,  p=^3  ft.  9  in. 

9.  e  =  2|  in.,  ^  =  4|  in.       12.  e  =  6  ft.  li  in.,^>  =  8  ft.  9|  in. 

Find  the  lateral  edges  of  the  prisms  whose  lateral  areas  and 
perimeters  of  right  sections  are  as  follows  : 

13.  /=187  sq.  in.,  j9=ll  in. 

14.  I  =  357  sq.  in.,  ^  =  21  in. 

15.  /  =169  sq.  in.,  p=litl  in. 

16.  The  lateral  surface  of  an  iron  bar  5  ft.  long  is  to  be 
gilded.  The  right  section  is  a  square  whose  area  is  2.89  sq.  in. 
How  many  square  inches  of  gilding  are  required  ? 

17.  A  right  prism  of  glass  is  2^  in.  long.  Its  right  section 
is  an  equilateral  triangle  whose  altitude  is  0.866  in.  (^  V3  in.). 
Find  the  lateral  surface. 

18.  Find  the  total  area  of  a  right  prism  whose  base  is  a  square 
with  area  5.29  sq.  in.,  and  whose  length  is  twice  its  thickness. 

19.  What  is  the  total  area  of  a  right  prism  whose  altitude 
is  32  in.,  and  whose  base  is  a  right  triangle  with  hypotenuse 
106  in.  and  with  one  side  84.8  in.? 

20.  Every  section  of  a  prism  made  by  a  plane  parallel  to  the 
lateral  edges  is  a  parallelogram. 


322 


BOOK  VII.    SOLID  GEOMETRY 


514.  Parallelepiped.    A  prism  whose  bases  are  parallelograms 
is  called  2,  jjavallelepixjed. 

The  word  is  also,  with  less  authority,  spelled  parallelopiped. 

515.  Right  Parallelepiped.    A  parallelepiped  whose  edges  are 
perpendicular  to  the  bases  is  called  a  right  parallelepiped. 

516.  Rectangular  Parallelepiped.   A  right  parallelepiped  whose 
bases  are  rectangles  is  called  a  rectangular  parallelepiped. 

By  §§  430  and  453  the  four  lateral  faces  are  also  rectangles. 


Rectangulai"  Parallelepiped 


Cube 


Oblique  Parallelepiped 


517.  Cube.  A  parallelepiped  whose  six  faces  are  all  squares 
is  called  a  cidje 

We  might  also  say  that  a  hexahedron  whose  six  faces  are  all  squares 
is  a  cube,  because  such  a  figure  would  necessarily  be  a  parallelepiped. 

518.  Unit  of  Volume.  In  measuring  volumes,  a  cube  whose 
edges  are  all  equal  to  the  unit  of  length  is  taken  as  the  unit 
of  volume. 

Thus,  if  •  we  are  measuring  the  contents  of  a  box  of  which  the  dimen- 
sions are  given  in  feet,  we  take  1  cubic  foot  as  the  unit  of  volume.  If  the 
dimensions  are  given  in  inches,  we  take  1  cubic  inch  as  the  unit. 

519.  Volume.  The  number  of  units  of  volume  contained  by 
a  solid  is  called  its  volume. 

520.  Equivalent  Solids.  If  two  solids  have  equal  volumes, 
they  are  said  to  be  e(piivalent. 

521.  Congruent  Solids.  If  two  geometric  solids  are  equal  in 
all  their  parts,  and  their  parts  are  similarly  arranged,  the  solids 
are  said  to  be  congruent. 


PARALLELEPIPEDS  323 

Proposition  III.    Theorem 

522.  Tivo  prisms  are  congruent  if  the  three  faces  which 
include  a  trihedral  angle  of  the  one  are  respectively  con- 
gruent to  three  faces  which  include  a  trihedral  angle  of 
the  other,  and  are  similarly  placed. 


Given  the  prisms  ^/and  A^I\  with  the  faces  AD^  AG^  -4/ re- 
spectively congruent  to  A^D\  A'G\  A'J\  and  similarly  placed. 

To  prove  that     AI  is  congruent  to  A' I'. 

Proof.  The  face  ABAE,  BAF,  EAF  are  equal  to  the  face 
A  B'A  'E',  BA  'F',  EA  '1^'  respectively.  §  142 

Therefore  the  trihedral  angles  A  and  A'  are  equal.  §  498 
Apply  the  trihedral  angle  A  to  its  equal  A'. 

Then  the  face  AD  coincides  with  A'D',  AG  with  A'G',  and 
A  J  with  A' J';  and  C  falls  at  C,  and  D  at  D'. 

The  lateral  edges  of  the  prisms  are  parallel.  §  446 

Therefore  CH  falls  along  C'H',  and  DI  along  D'l'.  §  94 

Since  the  points  F,  G,  and  /  coincide  with  F',  G',  and  J', 
each  to  each,  the  planes  of  the  upper  bases  coincide.  §  427 

Hence  H  coincides  with  H',  and  /  with  /'. 

Hence  the  prisms  coincide  and  are  congruent,  by  §  521.  Q.  e.  d. 

523.  Corollary  1.  Two  truncated  prisms  are  congruent 
under  the  conditions  given  in  Proposition  HI. 

524.  Corollary  2.  Two  right  prisms  having  congruent 
bases  and  equal  altitudes  are  congruent 


324  BOOK  VII.    SOLID  GEOMETRY 

Proposition  IV.    Theorem 

525.  An  oblique  prism  is  equivalent  to  a  right  prism 
whose  base  is  equal  to  a  right  section  of  the  oblique 
prism,  and  ivhose  altitude  is  equal  to  a  lateral  edge  of 
the  oblique  prism. 


Given  a  right  section  FI  of  the  oblique  prism  AD\  and  FV  a 
right  prism  whose  lateral  edges  are  equal  to  the  lateral  edges  of  AD^ . 

To  prove  that        AD'  is  equivalent  to  FI'. 

Proof.    If  from  the  equal  lateral  edges  of  AD'  and  FI'  we 
take  the  lateral  edges  of  FD',  which  are  common  to  both,  the 
remainders  AF  and  A'F',  BG  and  B'G',  etc.,  are  equal.       Ax.  2 
The  bases  FI  and  F'l'  are  congruent.  §  510 

Place  ^/  on  A' I'  so  that  FI  shall  coincide  with  FT. 
Then  FA,  GB,  etc.,  coincide  with  F'A',  G'B',  etc.  §  436 

Hence  the  faces  GA  and  G'A',  HB  and  II'B',  coincide. 
But  the  faces  FI  and  F'l'  coincide. 
.*.  the  truncated  prisms  A I  and  A' I'  are  congruent.    §  523 
.-.  AI-\-FD'  =  A'I'-{-FD'.  Ax.  1 

But  AI+FD'  =  AD', 

and  A  'I'  +  FD'  =  FI '.  Ax.  11 

Therefore        AD'  is  equivalent  to  FI',  by  Ax.  9.  q.e.d. 


PARALLELEPIPEDS  325 

Proposition  Y.    Theorem 


526.  The  opposite  faces  of  a  parallelepiped  are  con- 
gruent and  parallel. 


B 
Given  a  parallelepiped  ABCD-A'B'C'D^. 

To  prove  that  the  opposite  faces  AB'  and  DC  are  con- 
gruent and  parallel. 

Proof.                               AB  is  II  to  DC,  §  118 

and                                          AB  =  DC.  §125 
Likewise             yl^'  is  II  and  equal  to  DD\ 

.•.ZBAA'  =  ZCDD'.  §461 

.-.  AB'  is  11  to  DC.  §  461 

.-.  ^^'  is  congruent  to  DC,  by  §  132.  q.e.d. 

EXERCISE  85 

1.  If  in  the  above  figure  the  three  plane  angles  at  A  are 
80°,  70°,  75°,  what  are  all  the  other  angles  in  the  faces  ? 

2.  Given  a  parallelepiped  with  the  three  plane  angles  at 
one  of  the  vertices  85°,  75°,  60°,  to  iind  all  the  other  angles 
in  the  faces. 

3.  Given  a  rectangular  parallelepiped  lettered  as  in  the  fig- 
ure above,  and  with  AB  =  4:,  BC  =  S,  and  CC'  =  3|,  to  find  the 
length  of  the  diagonal  A  C'. 

4.  The  four  diagonals  of  a  rectangular  parallelepiped  are 
equal. 

5.  Compute  the  lengths  of  the  diagonals  of  a  rectangular 
parallelepiped  whose  edges  from  any  vertex  are  a,  h,  c. 


326 


BOOK  VII.    SOLID  GEOMETRY 


Proposition  YI.    Theorem 

527.  TJie  plane  passed  through  two  diagonally  oppo- 
site edges  of  a  parallelepiped  divides  the  parallelepiped 
into  two  equivalent  triangular  prisms. 


..H^nii^lilillin 

lihiiiiiiiiiiiiiiiiiiiiiffiiiiiiiiiii^ 

1 

D' 

^■■M\ 

m 

wJ^^^py 

lipffiliLrTr^-,^ 

^U^l^^^^c 

Given  the  plane  ACC'A'  passed  through  the  opposite  edges  AA^ 
and  CC  of  the  parallelepiped  AC^. 

To  prove  that  the  parallelepiped  AC  is  divided  into  two 
equivalent  triangular  prisms  ABC-B'  and  AC  D-D'. 

Proof.    Let  WXYZ  be  a  right  section  of  the  parallelepiped. 
The  opposite  faces  AB'  and  DC"  are  parallel  and  equal.   §  526 
Similarly,  the  faces  AD'  and  BC'  are  parallel  and  equal. 

.-.  WX  is  II  to  ZY,  and  WZ  to  XY.  §  453 

Therefore  WXYZ  is  a  parallelogram.  §  118 

The  plane  ACC'A'  cuts  this  parallelogram  WXYZ  in  the 

diagonal  WY.  §  429 

.-.A  WXY  is  congruent  to  AYZW.  §  126 

How  shall  it  be  proved  that  prism  ABC-B'  is  equivalent  to 
a  right  prism  with  base  WXY  and  altitude  AA'? 

How  shall  it  be  proved  that  prism  CDA-D'  is  equivalent  to 
a  right  prism  with  base  YZW  and  altitude  AA' ? 

How  are  these  two  right  prisms  known  to  be  equivalent  ? 

How  does  this  prove  the  proposition  ? 

Discussion.  What  is  the  corresponding  proposition  of  plane  geometry  ? 


PAKALLELEPIPEDS  327 

EXERCISE  86 

1.  The  lateral  faces  of  a  right  prism  are  rectangles. 

2.  The  diagonals  of  a  parallelepiped  bisect  one  another. 

3.  The  three  edges  of  the  trihedral  angle  at  one  of  the  ver- 
tices of  a  rectangular  parallelepiped  are  5  in.,  6  in.,  and  7  in. 
respectively.  Required  the  total  area  of  the  six  faces  of  the 
parallelepiped. 

4.  The  three  face  angles  at  one  vertex  of  a  parallelepiped 
are  each  60°,  and  the  three  edges  of  the  trihedral  angle  with 
that  vertex  are  3  in.,  2  in.,  1  in.  respectively.  Required  the 
total  area  of  the  six  faces.    Answer  to  two  decimal  places. 

5.  In  a  rectangular  parallelepiped  the  square  on  any  diag- 
onal is  equivalent  to  the  sum  of  the  squares  on  any  three  edges 
that  meet  at  one  of  the  vertices. 

6.  In  a  box  3  in.  deep  and  6  in.  wide  a  wire  1  ft.  long  can 
be  stretched  to  reach  from  one  corner  to  the  diagonally  oppo- 
site corner.  Required  the  length  of  the  box.  Answer  to  two 
decimal  places. 

7.  The  diagonal  of  the  base  of  a  rectangular  parallelepiped 
is  31f  in.  and  the  height  of  the  parallelepiped  is  23.7  in. 
Required  the  length  of  the  diagonal  of  the  parallelepiped. 

8.  The  total  area  of  the  six  faces  of  a  cube  is  18  sq.  in. 
Find  the  diagonal  of  the  cube. 

9.  The  diagonal  of  the  face  of  a  cube  equals  VIS.  Find 
the  diagonal  of  the  cube. 

10.  The  diagonal  of  a  cube  equals  2.75  V3.  Find  the  diagonal 
of  a  face  of  the  cube. 

11.  A  water  tank  is  3  ft.  long,  2  ft.  6  in.  wide,  and  1  ft.  9  in. 
deep.  How  many  square  feet  of  zinc  will  be  required  to  line 
the  four  sides  and  the  base,  allowing  1^  sq.  ft.  for  overlapping 
and  for  turning  the  top  edge  ? 


328 


BOOK  VII.    SOLID  GEOMETRY 
Proposition  VII.    Theorem 


528.    Tivo    rectangular   parallelepipeds    having    con- 
gruent bases  are  to  each  other  as  their  altitudes. 


Given  two  rectangular  parallelepipeds  P  andP',  with  congruent 
bases  and  with  altitudes  AB  and  A'B', 

To  prove  that  P  :  P'=  AB  :  A'B\ 

Case  1.    When  AB  and  A'B'  are  commensurable. 

Proof.  Suppose  a  common  measure  of  AB  and  A'B'  to  be 
contained  m  times  in  AB,  and  n  times  in  A'B'. 

Then  AB:A'B'  =  m:n. 

Apply  this  measure  to  AB  and  ^'J5',  and  through  the  several 
points  of  division  pass  planes  perpendicular  to  these  lines. 

These  planes  divide  the  parallelepiped  P  into  m  parallele- 
pipeds and  the  parallelepiped  P'  into  n  parallelepipeds,  con- 
gruent each  to  each.  §  524 

.-.  P:P' =  7n:n. 

.■.P:P'  =  AB:A'B',  by  Ax.  8.         q.e.d. 

The  proof  for  the  incommensurable  case  is  similar  to  that  in  other 
propositions  of  this  nature.  It  may  be  omitted  at  the  discretion  of  the 
teacher  without  destroying  the  sequence,  if  the  incommensurable  cases 
are  not  being  considered  by  the  class. 


PAEALLELEPIPEDS 
Case  2.    When  AB  and  AB'  are  incommensurable. 


329 


Proof.  Divide  A  B  into  any  number  of  equal  parts,  and  apply 
one  of  these  parts  to  A'B'  as  a  unit  of  measure  as  many  times 
as  J.  'i^'  will  contain  it. 

Since  ^i^  and  .4 '5'  are  incommensurable,  a  certain  number 
of  these  parts  will  extend  from  ^'  to  a  point  Z>,  leaving  a 
remainder  DB'  less  than  one  of  the  parts. 

Through  D  pass  a  plane  _L  to  A^B\  and  let  Q  denote  the 
parallelepiped  whose  base  is  the  same  as  that  of  P',  and  whose 
altitude  is  A  'D. 

Then  Q:  P  =  A'D:AB.  Case  1 

If  the  number  of  parts  into  which  AB  is  divided  is  indefi- 
nitely increased,  the  ratio  Q  :  P  approaches  P' :  P  as  a  limit, 
and  the  ratio  A'D:  AB  approaches  A'B' :  AB  SiS  Si  limit.       §  204 

The  remainder  of  the  proof  of  the  incommensurable  case 
is  substantially  as  in  the  proof  given  on  page  297,  and  it  is 
therefore  left  for  the  student. 

529.  Dimensions.  The  lengths  of  the  three  edges  of  a  rec- 
tangular parallelepiped  which  meet  at  a  common  vertex  are 
called  its  dijnensions. 

530.  Corollary.  Two  rectangular  parallelepipeds  wliich 
have  two  dimensions  in  common  are  to  each  other  as  their  third 
dimensions. 


330  BOOK  VIL    SOLID  GEOMETEY 

Proposition  YIII.    Theorem 

531.  Tivo  rectangular  parallelepipeds  having   equal 
altitudes  are  to  each  other  as  their  bases. 


Given  two  rectangular  parallelepipeds,  P  and  P',  and  a,  &,  c,  and 
a',  &',  c,  their  three  dimensions  respectively. 

F'~a'b'' 


To  prove  that 


Proof.  Let  Q  be  a  third  rectangular  parallelepiped  whose 
dimensions  are  a',  b,  and  c. 

Now  Q  has  the  two  dimensions  h  and  c  in  common  with  7', 
and  the  two  dimensions  a'  and  c  in  common  with  P'. 


Therefore 
and 


P      a 


Q 

~a'' 

Q 
P' 

h 

~  v' 

§530 


The  products  of  the  corresponding  members  of  these  two 
equations  give  p        ^^ 


}"   „v'^y^^-^- 


Q.E.D. 


532.  Corollary.  Two  rectangular  parallelepipeds  which 
have  one  dimension  in  common  are  to  each  other  as  the 
products  of  their  other  two  dimensions. 

For  any  edge  of  a  rectangular  parallelepiped  may  be  taken  as  the 
altitude,  whence  §  531  applies. 


PARALLELEPIPEDS 
PnoposiTioisr  IX.    Theorem 


331 


533.  Two  rectangular  parallelepipeds  are  to  each  other 
as  the  products  of  their  three  dimensions. 


Given  two  rectangular  parallelepipeds,  P  and  P',  and  a,  6,  c,  and 
a',  6',  c',  their  three  dimensions  respectively. 

P        abe 
F' 


To  prove  that 


,'hij 


Proof.    Let  Q  be  a  third  rectangular  parallelepiped  whose 
dimensions  are  a,  h\  and  c. 

Then  ''      ' 

and 


Q-V 

§530 

Q        ac 
P'      a'c' 

§532 

P        ahe     ,       .       _ 
P'  =  aVc"^^^'''^' 

Q.E.D. 

534.  Corollary  1.  The  volume  of  a  rectangular  parallele- 
pijyed  is  equal  to  the  product  of  its  three  dimensions. 

For  in  the  above  case,  \i  a'=¥=  c'=l,  then  P' =1x1x1=1  (§  518). 
But  the  volume  of  P  (§  519)  is  P  :  P',  and  P  :  P'=  a6c  :  1  (§  533).  There- 
fore the  volume  of  P  is  abc. 

535.  Corollary  2.  The  volume  of  a  rectangular  parallele- 
piped is  equal  to  the  product  of  its  base  and  altitude. 

For  the  volume  of  P  is  a6c,  and  ah  equals  the  base  and  c  the  altitude. 


332  BOOK  VII.    SOLID  GEOMETRY 

Proposition  X.    Theorem 

536.  The  volume  of  any  parallelepiped  is  equal  to  the 
product  of  its  base  hi/  its  altitude. 


Given  an  oblique  parallelepiped  P  of  volume  v,  with  no  two  of 
its  faces  perpendicular,  with  base  b  and  with  altitude  a. 

To  prove  that  v  =  ha. 

Proof.  Produce  the  edge  EF  and  the  edges  li  to  EF,  and  cut 
them  perpendicularly  by  two  parallel  planes  whose  distance 
apart  GI  is  equal  to  EF.  We  then  have  the  oblique  parallele- 
piped Q  whose  base  c  is  a  rectangle. 

Produce  the  edge  IK  and  the  edges  il  to  IK,  and  cut  them 
perpendicularly  by  two  planes  whose  distance  apart  AIJV  is 
equal  to  7A'.    We  then  have  the  rectangular  parallelepiped  R. 

Now  P  =  Q,  and  Q  =  R.  §  525 

.\P  =  R.  Ax.  8 

The  three  parallelepipeds  have  a  common  altitude  a.  §  455 

Also  b  =  c,  §323 

and  c  =  d.  §  133 

.\b  =  d.  Ax.  8 

But  the  volume  oi  R  =  da.  §  535 

Putting  P  for  R,  and  h  for  d,  we  have  v  =  ha,  by  Ax.  9.  q.e.d. 

537.  Corollary.  The  volume  of  auT/ parallelepiped  is  equal 
to  that  of  a  rectangular  parallelepiped  of  equivalent  hase  and 
equal  altitude. 


PARALLELEPIPEDS  333 

EXERCISE  87 

1.  Find  the  ratio  of  two  rectangular  parallelepipeds,  if  their 
dimensions  are  3,  4,  5,  and  9,  8,  10  respectively. 

2.  Eind  the  ratio  of  two  rectangular  parallelepipeds,  if  their 
altitudes  are  each  6  in.,  and  their  bases  5  in.  by  4  in.,  and  10  in. 
by  8  in.  respectively. 

3.  Find  the  volume  of  a  rectangular  parallelepiped  2  ft. 
6  in.  long,  1  ft.  8  in.  wide,  and  1  ft.  6  in.  high. 

4.  Find  the  volume  of  a  rectangular  parallelepiped  whose 
base  is  27  sq.  in.  and  whose  altitude  is  1^^  in. 

5.  The  volume  of  a  rectangular  parallelepiped  is  1152 
cu.  in.  and  the  area  of  the  base  is  half  a  square  foot.  Find 
the  altitude. 

6.  The  volume  of  a  rectangular  parallelepiped  with  a  square 
base  is  273.8  cu.  in.  and  the  altitude  is  5  in.  Find  the  dimen- 
sions. 

7.  A  rectangular  tank  full  of  water  is  7  ft.  3  in.  long  by 
4  ft.  6  in.  wide.  How  many  cubic  feet  of  water  must  be  drawn 
off  in  order  that  the  surface  may  be  lowered  a  foot  ? 

8.  Find  to  two  decimal  places  the  length  of  each  side  of  a 
cubic  reservoir  that  will  contain  exactly  a  gallon  (231  cu.  in.). 

9.  A  box  has  as  its  internal  dimensions  18  in.,  9^  in.,  and 
4^  in.  The  box  and  cover  are  made  of  steel  ^  in.  thick.  If  steel 
weighs  490  lb.  per  cubic  foot,  what  is  the  weight  of  the  box  ? 

10.  A  steel  rod  4  ft.  8  in.  long  is  2  in.  wide  and  1|-  in.  thick. 
How  much  does  it  weigh,  at  490  lb.  per  cubic  foot  ? 

11.  If  3  cu,  in.  of  gold  beaten  into  gold  leaf  will  cover 
75,000  sq.  in.  of  surface,  find  the  thickness  of  the  leaf. 

12.  The  sum  of  the  squares  on  the  four  diagonals  of  a  par- 
allelepiped is  equivalent  to  the  sum  of  the  squares  on  the 
twelve  edges. 


334  BOOK  VIL    SOLID  GEOMETRY 

Proposition  XL    Theorem 

538.  The  volume  of  a  triangular  jjrism  is  equal  to  the 
product  of  its  base  hy  its  altitude. 

D' 


Given  the  triangular  prism  ABC-B\  with  volume  v,  base  6,  and 
altitude  a. 

To  prove  that  v  =  ba. 

Proof.    Upon  the  edges  AB,  BC,  BB'  construct  the  parallele- 
piped ABCD-B'. 

Then  ABC-B'  =  ^  ABCD-B'.  §  527 

The  volume  of      ABCD-B'  =  ABCD  x  a.  §  536 

But  ABCD  ^  2  b.  §126 

.'.V  =  ^(2ba)  =  ha,hy  Ax.  9.  q.e.d. 

EXERCISE  88 

Find  the  volumes  of  the  triangular  prisms  whose  bases  and 
altitudes  are  as  follows  : 

1.  17  sq.  in.,  8  in. 

2.  15.75  sq.  ft.,  3  ft. 

3.  31  sq.  ft.,  1  ft.  8  in. 

4.  51  sq.  ft.,  2  ft.  9  in. 

5.  15.84  sq.  ft.,  3  ft.  10  in. 


6.  16f  sq.  in.,  2f  in. 

7.  221  sq.  in.,  4i  in. 

8.  331  sq.  in.,  7i  in. 

9.  42|  sq.  in.,  ^  in. 
10.  27f  sq.  in.,  3|  in. 


11.  12  sq.  ft.  75  sq.  in.,  2  ft.  7  in. 


PRISMS  335 

Proposition  XII.    Theorem 

539.  The  volume  of  any  prism  is  equal  to  the  product 
of  its  base  hy  its  altitude. 


Given  the  prism  AC^  with  volume  y,  base  6,  and  altitude  a. 

To  prove  that  v  =  ha. 

Proof.  It  is  possible  to  divide  any  prism  in  general  into 
what  kind  of  simpler  prisms  ? 

How  is  this  done  ? 

What  is  the  volume  of  each  of  these  simpler  prisms  (§  538)  ? 

What  is  the  sum  of  the  volumes  of  these  simpler  prisms  ? 

What  is  the  sum  of  their  bases  ? 

How  does  the  common  altitude  of  these  simpler  prisms 
compare  with  a,  the  altitude  of  the  given  prism  ? 

What  conclusion  can  be  drawn  from  these  statements  ? 

Write  the  proof  in  full. 

540.  Corollary  1.  Prisms  having  equivalent  bases  are  to 
each  other  as  their  altitudes ;  prisms  having  equal  altitudes 
are  to  each  other  as  their  bases. 

Write  the  proof  in  full. 

541.  Corollary  2.  Prisms  having  equivalent  bases  and 
equal  altitudes  are  equivalent. 

Write  the  proof  in  full. 


336  BOOK  VII.    SOLID  GEOMETRY 

EXERCISE  89 

1.  If  the  length  of  a  rectangular  parallelepiped  is  18  in.,  the 
width  9  in.,  and  the  height  8  in.,  find  the  total  area  of  the  surface. 

2.  Find  the  volume  of  a  triangular  prism,  if  its  height  is 
15  in.  and  the  sides  of  the  base  are  6  in.,  5  in.,  and  5  in. 

3.  Find  the  volume  of  a  prism  whose  height  is  15  ft.,  if 
each  side  of  the  triangular  base  is  10  in, 

4.  The  base  of  a  right  prism  is  a  rhombus  of  which  one 
side  is  20  in.,  and  the  shorter  diagonal  24  in.  The  height  of 
the  prism  is  30  in.    Find  the  entire  surface  and  the  volume. 

5.  How  many  square  feet  of  lead  will  be  required  to  line  an 
open  cistern  which  is  4  ft.  6  in.  long,  2  ft.  8  in.  wide,  and  con- 
tains 42  cu.  ft.? 

6.  An  open  cistern  6  ft.  long  and  41  ft.  wide  holds  108 
cu.  ft.  of  water.  How  many  square  feet  of  lead  will  it  take 
to  line  the  sides  and  bottom  ? 

7.  One  edge  of  a  cube  is  e.  Find  in  terms  of  e  the  surface, 
the  volume,  and  the  length  of  a  diagonal  of  the  cube. 

8.  The  diagonal  of  one  of  the  faces  of  a  cube  is  d.  Find  in 
terms  of  d  the  volume  of  the  cube. 

9.  The  three  dimensions  of  a  rectangular  parallelepiped  are 
a,  h,  c.  Find  in  terms  of  a,  h,  and  c  the  volume  and  the  area  of 
the  surface. 

10.  Find  the  volume  of  a  prism  with  bases  regular  hexagons, 
if  the  height  is  10  ft.  and  each  side  of  the  hexagons  is  10  in. 

11.  An  open  cistern  is  made  of  iron  ^  in.  thick.  The  inner 
dimensions  are :  length,  4  ft.  6  in. ;  breadth,  3  ft. ;  depth,  2  ft. 
6  in.  What  will  the  cistern  weigh  when  empty  ?  when  full  of 
water  ?  (A  cubic  foot  of  water  weighs  62^  lb.  Iron  is  7.2  times 
as  heavy  as  water ;  that  is,  the  specific  gravity  of  iron  is  7.2.) 


PYRAMIDS  337 

542.  Pyramid.  A  polyhedron  of  which  one  face,  called  the 
hasBj  is  a  polygon  of  any  number  of  sides  and  the  other  faces 
are  triangles  having  a  common 

vertex  is  called  2i  pyramid. 

The   triangular  faces  having  a 
common  vertex  are  called  the  lateral 

faces,  their  intersections  are  called  /    gg    f/g  /Milium 

the  lateral  edges,  and  their  common 
vertex  is  called  the  vertex  of  the 
pyramid.  The  base  of  a  pyramid 
may  be  any  kind  of  a  polygon,  but 
usually  a  convex  polygon  is  taken. 

543.  Lateral  Area.  The  sum  of  the  areas  of  the  lateral  faces 
of  a  pyramid  is  called  the  lateral  area  of  the  pyramid. 

544.  Altitude.  The  perpendicular  distance  from  the  vertex 
to  the  plane  of  the  base  is  called  the  altitude  of  the  pyramid. 

545.  Pyramids  classified  as  to  Bases.  Pyramids  are  said  to 
be  triangular,  quadrangular,,  and  so  on,  according  as  their 
bases  are  triangles,  quadrilaterals,  and  so  on, 

A  triangular  pyramid  has  four  triangular  faces  and  is  called  a  tetra- 
hedron.  Any  one  of  its  faces  may  be  taken  as  the  base. 

546.  Regular  Pyramid.  If  the  base 
of  a  pyramid  is  a  regular  polygon 
whose  center  coincides  with  the  foot 
of  the  perpendicular  let  fall  from  the 
vertex  to  the  base,  the  pyramid  is 
QdW^di  di,  regular  pyramid.  ^  ^.  ji 

A  regular  pyramid  is  also  called  a  right 
'pyramid. 

547.  Slant  Height  of  a  Regular  Pyramid.  The  altitude  of 
any  one  of  the  lateral  faces  of  a  regular  pyramid,  drawn 
from  the  vertex  of  the  pyramid,  is  called  the  slant  height. 

The  slant  height  is  the  same  whatever  face  is  taken  (§  439).  Only  a 
regular  pyramid  can  have  a  slant  height. 


338 


BOOK  VII.    SOLID  GEOMETRY 


548.  Properties  of  Regular  Pyramids.  Among  the  properties 
of  regular  pyramids  the  following  are  too  evident  to  require 
further  proof  than  that  referred  to  below : 

(1)  The  lateral  edges  of  a  regular  pyramid  are 
equal  (§  439). 

(2)  The  lateral  faces  of  a  regular  pyramid  are 
congruent  isosceles  triangles  (§  80). 

(3)  The  slant  height  of  a  regular  pyramid  is 
the  same  for  all  the  lateral  faces  (§  439). 

549.  Frustum  of  a  Pyramid.  The  portion  of  a  pyramid  in- 
cluded between  the  base  and  a  section  parallel  to  the  base  is 
called  a  frustum^  of 
a  pyramid. 

The  base  of  the  pyra- 
mid and  the  parallel 
section  are  called  the 
6ases  of  the  frustum. 

A  tnore  general  term, 
including  frustum  as  a  special  case,  is  truncated  pyramid,  the  portion  of 
a  pyramid  included  between  the  base  and  any  section  made  by  a  plane 
that  cuts  all  the  lateral  edges.   This  term  is  little  used. 

550.  Altitude  of  a  Frustum.  The  perpendicular  distance 
between  the  bases  is  called  the  altitude  of  the  frustum. 

E.g.  C^C  is  the  altitude  of  the  frustum  in  the  above  figure. 

551.  Lateral  Faces  of  a  Frustum.  The  portions  of  the  lateral 
faces  of  a  pyramid  that  lie  between  the  bases  of  a  frustum  are 
called  the  lateral  faces  of  the  frustum. 

In  the  case  of  a  frustum  of  a  regular  pyramid  the  lateral  faces  are 
congruent  isosceles  trapezoids.  The  sum  of  the  areas  of  the  lateral  faces 
is  called  the  lateral  area  of  the  frustum. 

552.  Slant  Height  of  a  Frustum.  The  altitude  of  one  of  the 
trapezoid  faces  of  a  frustum  of  a  regular  pyramid  is  called  the 
slant  height  of  the  frustum. 

Thus  MM'  in  the  above  fi2:ure  is  the  slant  heisrht. 


PYRAMIDS  339 

Proposition  XIII.    Theorem 

553.  The  lateral  area  of  a  regular  pyramid  is  equal 
to  half  the  product  of  its  slant  height  by  the  perimeter 
of  its  base. 


Given  the  regular  pyramid  V-ABCDE^  with  /  the  lateral  area, 
s  the  slant  height,  and  p  the  perimeter  of  the  base. 


To  prove  that 


l  =  h 


Proof.     The   A  VAB,  VBC,  VCD,  VDE,  and  VEA   are    con^ 

gruent.  §  548 

The  area  of  each  A  =  ^  s  x  its  base. 

The  sum  of  the  bases  of  the  triangles  =p. 

.  • .  the  sum  of  the  areas  of  these  A  =  ^  sp. 

But  the  sum  of  the  areas  of  these  A  =  I. 

.\l  =  ^sjj,  by  Ax.  8. 

554.  Corollary.  The  lateral  area 
of  the  frustum  of  a  regular  pyramid 
is  equal  to  half  the  sum  of  the  perim- 
eters of  the  bases  multiplied  by  the 
slant  height  of  the  frustum. 

How  is  the  area  of  a  trapezoid  found  (§  329)  ?  Are  these  trapezoids 
congruent  ?  What  is  the  sum  of  their  lower  bases  ?  of  their  upper  bases  ? 
What  is  the  sum  of  their  areas  ?    Insert  the  formula. 


340  BOOK  VIL    SOLID  GEOMETRY 

Proposition  XIV.    Theorem 

555.  If  a  pyramid  is  cut  hy  a  jjlcme  parallel  to  the 
base  :   - 

1.  77ie  edges  and  altitude  are  divided  proportionally , 

2.  I'he  section  is  a  polygon  similar  to  the  base. 

V 


1  ^mf'  \ 

'  i  \  \  \ 

/     4-k\^n     \ 

IB                   V                  \ 

Given  the  pyramid  V-ABCDE  cut  by  a  plane  parallel  to  its  base, 
intersecting  the  lateral  edges  in  A\  B\  C\  D\  E\  and  the  alti- 
tude FO  in  O'. 

,     ^  ,        VA'      VB'  VO' 

1.  To  prove  that   yj  =  -^  =  •  •  •  =  —  • 

Proof.     Since  the  plane  A^D^  is  II  to  the  plane  AD,        Given 

.-.A'B'  is  II  to  J.^,  5'C'is  II  to  5C,...,  and  ^'0' is  II  to  .4  0.  §  453 

VA^      VB'  VO'  -      ^  ^^, 

.-. = =  ...  = ,  by  §  274.  Q.E.D. 

VA        VB  VO      ^ 

2.  To  prove  the  section  A' B' C D' E'  similar  to  the  base  A  B  CDE. 

Proof.  Since  AVA'B'  is  similar  to  AVAB,  AVB'C'  similar 
to  AVBC,  and  so  on  (why  ?),  how  can  the  corresponding  sides 
of  the  polygons  be  proved  proportional  ? 

Since  A'B'  is  II  to  AB,  B'C  to  BC,  etc.  (why  ?), 

how  can  the  corresponding  angles  be  proved  equal  ? 

Then  why  is  A'B'C'D'E'  similar, to  ABCDE? 


PYRAMIDS 


341 


556.  Corollary  1.  Any  section  of  a  pyramid  parallel  to 
the  base  is  to  the  base  as  the  square  of  the  distance  from  the 
vertex  is  to  the  square  of  the  altitude  of  the  pyramid. 


§655 


For 

VO' 
VO 

VA' 
VA 
A'B' 
AB 

Therefore 

But,  from  similar 

polygons. 

A'B'^ 
AB' 

A'B'C'B'E' 

A'B'' 

ABCDE 
Hence,  by  substituting, 

A'B'C'B'E'  _ 

AB' 

v(y^ 

§288 


§270 


§334 


Ax.  8 


ABCDE         YQ^ 


557.  Corollary  2.  If  tivo  pyramids  have  equal  altitudes 
and  equivalent  bases,  sections  made  by  planes  parallel  to  the 
bases,  and  at  equal  distances  fro7n  the  vertices,  are  equivalent. 

What  is  the  ratio  of  A'B'C'D'E'  to  ABC  BE  ? 
How  can  this  be  shown  to  equal  VO'  :  VO  ? 
What  is  the  ratio  of  X'TZ'  to  XYZ  ? 
How  can  this  be  shown  to  equal  WP'  :  WP  ? 
Are  the  ratios  VO'^  :  VO'  and  WP'^  :WP'  equal  ? 
Since  it  is  given  that  ABCBE  =  XYZ,  what  can  be  said  of  A'B'C'B'E' 
and  X'TZ'  ? 


342  BOOK  VIL    SOLID  GEOMETRY 

Peoposition  XV.    Theorem 

558.  Tico    triangular   pyramids    having    equivalent 
bases  and  equal  altitudes  are  equivalent. 


having 


B  B' 

Given  two  triangular  pyramids,  V-ABC  and  F'-A'5'C', 
equivalent  bases  and  equal  altitudes. 

To  prove  that  V-ABC  and  V'-A'B'C  are  equivalent. 
Proof.    Suppose  the  pyramids  are  not  equivalent,  and 
V'-A'B'OV-ABC. 

Place  the  bases  in  the  same  plane,  and  suppose  the  altitude 
divided  into  n  equal  parts,  calling  each  of  these  parts  h. 

Through  the  points  of  division  pass  planes  parallel  to  the 
base,  cutting  the  pyramids  in  DBF,  GHI,  •  •  • ,  D'E'F',  G'H'I', .... 

On  A'B'C,  D'E'F',  G'H'I',  and  other  parallel  sections,  if  any, 
construct  prisms  with  lateral  edges  parallel  to  A'V',  and  with 
altitude  h.    In  the  figure  these  are  represented  by  X',  Y',  and  Z'. 

On  DBF,  GHI,  and  other  parallel  sections,  if  any,  as  upper 
bases,  construct  the  prisms  Y,  Z,  with  lateral  edges  parallel  to 
VA,  and  with  altitude  h. 

Then  since  DBF  =  D'E'F',  §  557 

and  h  =  h,  Iden. 

.*.  prism  Y=  prism  Y'.  §  541 
Similarly                     prism  Z  =  prism  Z'. 


PYKAMIDS  343 

But  X'-^Y'  +  Z'>  V'-A  'B'C, 

and  Y-^Z<  V-ABC.  Ax.  11 

.-.  V'-A'B^C  -  V-ABC  <  X'  -f-  r  H-  Z'  -  (  Y+  Z), 
or  V'-A '^'C  -  V-ABC  <  X'. 

That  is,  the  difference  between  the  pyramids  must  be  less 
than  the  difference  between  the  sets  of  prisms. 

Now  by  increasing  n  indefinitely,  and  consequently  de- 
creasing h  indefinitely,  X'  can  be  made  less  than  any  assigned 
quantity. 

Hence  whatever  difference  we  suppose  to  exist  between  the 
pyramids,  X'  can  be  made  smaller  than  that  supposed  difference. 

But  this  is  absurd,  since  we  have  shown  that  X'  is  greater 
than  the  difference,  if  any  exists. 

Hence  it  leads  to  a  manifest  absurdity  to  suppose  that 
V'-A'B'O  V-ABC. 

In  the  same  way  it  leads  to  an  absurdity  to  suppose  that 
V-ABC>  V'-A'B'C. 

.'.  V-ABC  =  V'-A'B'C.  Q.E.D. 

EXERCISE  90 

1.  The  slant  height  of  a  regular  pyramid  is  6  in.,  and  the 
base  is  an  equilateral  triangle  of  altitude  2  V 3  in.  Find  the 
lateral  area  of  the  pyramid. 

2.  The  slant  height  of  a  regular  triangular  pyramid  equals 
the  altitude  of  the  base.  The  area  of  the  base  is  Vs  sq.  ft. 
Find  the  total  area  of  the  pyramid. 

3.  A  pyramid  has  for  its  base  a  right  triangle  with  hy- 
potenuse 5  and  shortest  side  3.  Another  one  of  equal  altitude 
has  for  its  base  an  equilateral  triangle  with  side  2  v  2  Vs, 
Prove  the  pyramids  equivalent. 


344  BOOK  VII.    SOLID  GEOMETRY 

Proposition  XVI.    Theorem 

559.  The  volume  of  a  triangular  pyramid  is  equal  to 
one  third  the  product  of  its  base  hy  its  altitude. 


Given  the  triangular  pyramid  E-ABC^  with  volume  v,  base  &, 
and  altitude  a. 

To  prove  that  v  =  ^ha. 

Proof.    On  the  base  ABC  construct  a  prism  ABC-DEF. 

Through  DE  and  EC  pass  a  plane  CDE. 
Then  the  prism  is  composed  of  three  triangular  pyramids 
E-ABC,  E-CFD,  and  E-ACD. 

Now  the  pyramids  E-CFD  and  E-A  CD  have  the  same  altitude 
and  equal  bases  CFD  and  A  CD.  §  126 

.  • .  E-CFD  =  E-A  CD.  §.  558 

But  pyramid  E-CFD  is  the  same  as  pyramid  C-DEF, 
which  has  the  same  altitude  as  pyramid  E-ABC, 

and  has  base  DEF  equal  to  base  ABC.  §  511 

.-.  E-CFD  =  E-ABC.  §  558 

.'.  E-ABC  =  E-CFD  =  E-A  CD.  Ax.  8 

.*.  pyramid  E-ABC  =  i  prism  ABC-DEF. 

But  the  volume  of  ABC-DEF  =  ha.  §  539 

.-.  V  =  1^  ^a,  by  Ax.  4.  Q.E.D. 

560.  Corollary.  The  volume  of  a  triangular  pyramid  is 
equal  to  one  third  the  volume  of  a  triangular  prism  of  the 
same  base  and  altitude. 


-       PYRAMIDS 
Proposition  XVII.    Theorem 


345 


561.  The  volume  of  any  2^yfamid  is  equal  to  one  third 
the  product  of  its  base  hy  its  altitude. 


Given  the  pyramid  V-ABCDEy  with  volume  y,  base  &,  and  alti- 
tude a. 

To  prove  that  v  =  ^  ha. 

Proof.  Through  the  edge  VD  and  the  diagonals  of  the  base, 
DA,  DB,  pass  planes. 

These  planes  divide  the  pyramid  V- ABODE  into  three  tri- 
angular pyramids. 

What  can  be  said  as  to  the  altitudes  of  the  original  pyramid 
and  of  the  triangular  pyramids  ? 

What  can  be  said  as  to  the  base  of  the  original  pyramid  in 
relation  to  the  bases  of  the  triangular  pyramids  ? 

What  is  the  volume  of  each  triangular  pyramid  ? 

What  is  the  sum  of  the  volumes  of  the  triangular  pyramids  ? 

Complete  the  proof. 

562.  Corollary.  The  volumes  of  two  pyramids  are  to  each 
other  as  the  products  of  their  bases  and  altitudes  ;  pyramids 
having  equivalent  bases  are  to  each  other  as  their  altitudes; 
pyramids  having  equal  altitudes  are  to  each  other  as  their 
bases;  pyramids  having  equivalent  bases  and  equal  altitudes 
are  equivalent. 


346  BOOK  VII.    SOLID  GEOMETRY 

EXERCISE  91 

Find  the  lateral  areas  of  regular  pyramids^  given  the  slant 
heights  and  the  perimeters  of  the  bases,  as  follows  : 

1.  s  =  34  in.,  j9  =  57  in.        3.  s  =  2  ft.  7  m.,p  =  4  ft.  6  in. 

2.  s  =  ^  \n.,p  =  ll\  in.     4.  s  =  127  ft.  5  in.,j9  =  63ft.2in. 

Find  the  lateral  areas  of  frustums  of  regular  pyramids^ 
given  the  slant  heights  of  the  frustums  and  the  perimeters  of 

the  bases,  as  follows  : 

5.  s  =  4:  in.,  p  =  S  in.,  p'  =  6  in. 

6.  8=5^  in.,  j9  =  9|  in.,  p'  =  7|  in. 

7.  s  =  2  ft.  3  in.,  j9  =  4  ft.  8  in.,  p'  =  3  ft.  9  in. 

Find  the  volumes  of  pyramids,  given  the  altitudes  and  the 
areas  of  the  bases,  as  follows  : 

S.  a  =  7  in.,  ^  =  9  sq.  in.       11.  a  =  S^  in.,  ^  =  5^  sq.  in. 
9.  a  =  6  in.,  b  =  2S  sq.  in.      12.  a  =  4^  in.,  b  =  19  sq.  in. 
10.  a  =  17  in.,  ^  =  51  sq.  in.      13.  a  =  27.5  ft.,  b  =  325  sq.  ft. 

Find  the  lateral  areas  of  regular  pyramids,  given  the  slant 
heights,  the  number  of  sides  of  the  bases,  and  the  length  of 
each  side,  as  follows  : 

14.  s  =  2.3  in.,  n  =  4.,l=  2.1  in. 

15.  s  =  3.7  in.,  n  =  6,l=  2.9  in. 

16.  s  =  5.33  in.,  n  =  S,l=S  in. 

Find  the  volumes  of  pyramids,  given  the  altitudes  and  a 
description  of  the  bases,  as  follows  : 

17.  a  =  7  in.,  the  base  a  square  with  side  2  in. 

18.  a  =  6|  in.,  the  base  a  square  with  diagonal  3  V2  in. 

19.  a  =  8.9  in.,  the  base  a  triangle  with  each  side  3.7  in. 


PYRAMIDS  347 

20.  Find  the  lateral  area  of  a  regular  pyramid,  if  the  slant 
height  is  16  ft.  and  the  base  is  a  hexagon  with  side  12  ft. 

21.  Find  the  lateral  area  of  a  regular  pyramid,  if  the  slant 
height  is  8  ft.  and  the  base  is  a  pentagon  with  side  5  ft. 

22.  Find  the  total  surface  of  a  regular  pyramid,  if  the  slant 
height  is  6  ft.  and  the  base  is  a  square  with  side  4  ft. 

23.  Find  the  total  surface  of  a  regular  pyramid,  if  the  slant 
height  is  18  ft.  and  the  base  is  a  square  with  side  8  ft. 

24.  Find  the  total  surface  of  a  regular  pyramid,  if  the  slant 
height  is  16  ft.  and  the  base  is  a  triangle  with  side  8  ft. 

25.  The  volume  of  a  pyramid  is  26  cu.  ft.  936  cu.  in.  and 
each  side  of  its  square  base  is  3  ft.  6  in.    Find  the  height. 

26.  The  volume  of  a  pyramid  is  20  cu.  ft.  and  the  sides  of 
its  triangular  base  are  5  ft.,  4  ft.,  and  3  ft.  respectively.  Find 
the  height. 

27.  Find  the  volume  of  a  regular  pyramid  with  a  square 
base  whose  side  is  40  ft.,  the  lateral  edge  being  101  ft. 

28.  Find  the  volume  of  a  regular  pyramid  whose  slant  height 
is  12  ft.  and  whose  base  is  an  equilateral  triangle  inscribed  in 
a  circle  of  radius  10  ft. 

29.  Having  given  the  base  edge  a  and  the  total  surface  t  of 
a  regular  pyramid  with  a  square  base,  find  the  height  h. 

30.  Having  given  the  base  edge  a  and  the  total  surface  t  of 
a  regular  pyramid  with  a  square  base,  find  the  volume  v. 

31.  The  eight  edges  of  a  regular  pyramid  with  a  square  base 
are  equal  and  the  total  surface  is  t.    Find  the  edge. 

32.  Find  the  base  edge  «-  of  a  regular  pyramid  with  a  square 
base,  having  given  the  height  h  and  the  total  surface  t. 

33.  Show  how  to  find  the  volume  of  any  polyhedron  by 
dividing  the  polyhedron  into  pyramids.  ^ 


348  BOOK  VII.    SOLID  GEOMETRY 

Proposition  XVIII.    Theorem 

563.  The  frustum  of  a  triangular  pyramid  is  equiva- 
lent to  the  sum  of  three  j^yramids  whose  commoii  altitude 
is  the  altitude  of  the  frustum  and  whose  bases  are  the 
lower  base,  the  upper  base,  and  the  mean  proportional 
between  the  tivo  bases  of  the  frustum. 


Given  the  frustum  of  a  triangular  pyramid,  ABC-DEF,  having 
ABC^  or  &,  for  its  lower  base ;  DEFy  or  &',  for  its  upper  base ;  and 
the  altitude  a. 

To  prove  that   ABC-DEF  =::\ah -\- \aV  +  \a  ^hh'. 

Proof.  Through  A,  E,  and  C,  and  also  through  C,  D,  and  E, 
pass  planes  dividing  the  frustum  into  three  pyramids. 

Then  E-ABC  =  ^ab, 

and  C-DEF=i-ab'.  §559 

It  therefore  remains  only  to  prove  that  E-A  CD  =  ^  a^/W. 
We  see  by  the  iigure  that  we  may  speak  of  E-ABC  as  C-ABE, 
and  of  E-A  CD  as  C-AED. 

But  C-ABE  :  C-AED=  A  ABE  :  A  AED.  §  562 

Since  A  ABE  and  AED  have  for  a  common  altitude  the 
altitude  of  the  trapezoid  ABED, 

.-.A  ABE:  A  AED  =  AB  :  DE.  §  327 

.-.  C-ABE  :  C-AED  =  AB:  DE,  Ax.  8 

or                        E-ABC  :  E-A  CD  =  AB  :  DE.  Ax.  9 


PYEAMIDS  349 

In  like  manner  E-ACD  and  E-CFD  have  a  common  vertex 
E  and  have  their  bases  in  the  same  plane,  A  CFD,  so  that 

E-A  CD  :  E-CFD  =  AACD:A  CFD.  %  562 

Since  A  A  CD  and  CFD  have  for  a  common  altitude  the  alti- 
tude of  the  trapezoid  A  CFD, 

.-.AACD-.A  CFD  =  AC  :DF.  §  327 

.  • .  E-A  CD  :  E-CFD  =  AC  :  DF.  Ax.  8 

But  A  DEF  is  similar  to  A  ABC.  §  555 

.-.  AB  :  DE  =  AC  :  DF.  §282 

.-.  E-ABC  :  E-A CD  =  AC  :  DF.  Ax.  8 

.  • .  E-ABC  :  E-A  CD  =  E-A  CD  :  E-CFD.  Ax.  8 

But  E-CFD  is  the  same  as  C-DEF,  which  has  been  shown  to 

equal  ^  ab'. 

.'.^ah:  E-A  CD  =  E-A  CD  :  ^  ab'.  Ax.  9 

.-.  E-ACD  =  -V^abxi ab'  §  262 

=  ia  Vbb'. 

.'.  E-ABC  +  C-DEF  +  E-A CD  =  iab  +  i ab'-^^aVbb'.      Ax.  1 

That  is,  ABC-DEE  =  ^ab -\- ^ab' -{- ^a  Vbb',  by  Ax.  9.    Q. e. d. 

564.  Corollary  1.  The  volume  of  a  frustum  of  a  tri- 
angular pyramid  may  he  expressed  as  ^  a  (h  -\- h'  -{-  VW). 

For  we  may  factor  by  i  a. 

565.  Corollary  2.    The    volume    of  a  frustum    of  any 

pyramid  is  equal  to  the  sum  of  the  volumes  of  three  pyramids 

whose  common  altitude  is  the  altitude  of  the  frustum,  and 

whose  bases  are  the  lower  base,  the  upper  base,  and  the  inean 

proportional  between  the  bases  of  the  frustum. 

For  any  pyramid  may  be  divided  into  triangular  pyramids,  and  this 
separates  any  of  its  frustums  into  frustums  of  triangular  pyramids. 
These  pyramids  have  a  common  altitude,  and  the  frustums  also  have  a 
common  altitude,  so  that  we  have  only  to  find  the  volumes  of  the  frus- 
tums by  §  563,  and  then  add  them. 


350 


BOOK  VII.    SOLID  GEOMETRY 


566.  Polyhedrons  classified  as  to  Faces.  A  polyhedron  of 
four  faces  is  called  a  tetrahedron ;  one  of  six  faces,  a  hexahe- 
dron ;  one  of  eight  faces,  an  octahedron ;  one  of  twelve  faces, 
a  dodecahedron  ;  one  of  twenty  faces,  an  icosahedron. 


Tetrahedron      Hexahedron       Octahedron      Dodecahedron      Icosahedron 

567.  Regular  Polyhedron.  A  polyhedron  whose  faces  are  con- 
gruent regular  polygons,  and  whose  polyhedral  angles  are  equal, 
is  called  a  regular  polyhedron. 

It  is  proved  on  page  351  that  it  is  possible  to  have  only  five  regular 
polyhedrons.     They  may  be  constructed  from  paper  as  follows  : 


Draw  on  stiff  paper  the  diagrams  given  above.  Cut  through  the  full 
lines  and  paste  strips  of  paper  on  the  edges  as  shown.  Fold  on  the  dotted 
lines,  and  keep  the  edges  in  contact  by  the  pasted  strips  of  paper. 


EEGULAR  POLYHEDRONS  351 

Proposition  XIX.    Problem 

568.  To  determine  the  number  of  regular  convex  poly- 
hedrons possible. 

A  convex  polyhedral  angle  must  have  at  least  three  faces, 
and  the  sum  of  its  face  angles  must  be  less  than  360°  (§  495). 

1.  Since  each  angle  of  an  equilateral  triangle  is  60°,  convex 
polyhedral  angles  may  be  formed  by  combining  three,  four,  or 
five  equilateral  triangles.  The  sum  of  six  such  angles  is  360°, 
and  therefore  is  greater  than  the  sum  of  the  face  angles  of  a 
convex  polyhedral  angle.  Hence  three  regular  convex  polyhe- 
drons are  possible  with  equilateral  triangles  for  faces. 

2.  Since  each  angle  of  a  square  is  90°,  a  convex  polyhedral 
angle  may  be  formed  by  combining  three  squares.  The  sum  of 
four  such  angles  is  360°,  and  therefore  is  greater  than  the  sum 
of  the  face  angles  of  a  convex  polyhedral  angle.  Hence  one 
regular  convex  polyhedron  is  possible  with  squares. 

3.  Since  each  angle  of  a  regular  pentagon  is  108°  (§  145),  a 
convex  polyhedral  angle  may  be  formed  by  combining  three 
regular  pentagons.  The  sum  of  four  such  angles  is  432°,  and 
therefore  is  greater  than  the  sum  of  the  face  angles  of  a  convex 
polyhedral  angle.  Hence  one  regular  convex  polyhedron  is 
possible  with  regular  pentagons. 

4.  The  sum  of  three  angles  of  a  regular  hexagon  is  360°,  of 
a  regular  heptagon  is  greater  than  360°,  and  so  on. 

Hence  only  five  regular  convex  polyhedrons  are  possible. 

The  regular  polyhedrons  are  the  regular  tetrahedron,  the 
regular  hexahedron,  or  cube,  the  regular  octahedron,  the  regular 
dodecahedron,  and  the  regular  icosahedron.  q.e.f. 

It  adds  greatly  to  a  clear  understanding  of  the  five  regular  poly- 
hedrons if  they  are  constructed  from  paper  as  suggested  in  §  567. 

Since  these  solids  were  extensively  studied  by  the  pupils  of  Plato,  the 
great  Greek  philosopher,  they  are  often  called  the  Platonic  Bodies. 


352  BOOK  VII.    SOLID  GEOMETRY 

EXERCISE  92 

Find  the  volumes  of  frustums  of  pyramids^  the  altitudes  and 
the  bases  of  the  frustums  being  given,  as  follows  : 

1.  a  =  3  in.,  b  =  S  sq.  in.,  b'  =  2  sq.  in. 

2.  a  =  4:^  in.,  ^  =  8^  sq.  in.,  b'  =  3  sq.  in. 

3.  a  =  3.2  in.,  b  =  2  sq.  in.,  b'  =  0.18  sq.  in. 

4.  a  =  2  ft.  6  in.,  ^»  =  10  sq.  ft.,  b'  =  2  sq.  ft.  72  sq.  in. 

5.  a  =  3  ft.  7  in.,  5  -=  24  sq.  ft.  72  sq.  in.,  b'  =2  sq.  ft. 

6.  A  pyramid  2  in.  high,  with  a  base  whose  area  is  8  sq.  in., 
is  cut  by  a  plane  parallel  to  the  base  1  in.  from  the  vertex. 
Eind  the  volume  of  the  frustum. 

7.  A  pyramid  3  in.  high,  with  a  base  whose  area  is  81  sq.  in., 
is  cut  by  a  plane  parallel  to  the  base  2  in.  from  the  base.  Find 
the  volume  of  the  frustum. 

8.  The  lower  base  of  a  frustum  of  a  pyramid  is  a  square 
4  in.  on  a  side.  The  side  of  the  upper  base  is  half  that  of  the 
lower  base,  and  the  altitude  of  the  frustum  is  the  same  as  the 
side  of  the  upper  base.    Find  the  volume  of  the  frustum. 

9.  The  lower  base  of  a  frustum  of  a  pyramid  is  a  square 
3  in.  on  a  side.  The  area  of  the  upper  base  is  half  that  of  the 
lower  base,  and  the  altitude  of  the  frustum  is  2  in.  Find  to 
two  decimal  places  the  volume  of  the  frustum. 

10.  A  pyramid  has  six  edges,  each  1  in.  long.  Find  to  two 
decimal  places  the  volume  of  the  pyramid. 

11.  A  regular  tetrahedron  has  a  volume  2  V2  cu.  in.  Find  to 
two  decimal  places  the  length  of  an  edge. 

12.  The  base  of  a  regular  pyramid  is  a  square  I  ft.  on  a  side. 
The  slant  height  is  s  ft.    Find  the^rea  of  the  entire  surface. 

13.  Consider  the  formula  v  =  ^^  a  (b -{-b' +  Vbb'),  of  §  564, 
when  b'  =  0.  Discuss  the  meaning  of  the  result.  Also  discuss 
the  case  in  which  b  =  b'. 


CYLINDERS 


353 


569.  Cylindric  Surface.  A  surface  generated  by  a  straight 
line  which  is  constantly  parallel  to  a  fixed  straight  line,  and 
touches  a  fixed   curve   not   in 

the  plane  of  the  straight  line, 
is  called  a  cylindric  surface^  or 
a  cylindrical  surface. 

The  moving  line  is  called  the 
generatrix  and  the  fixed  curve  the 
directrix.  In  the  figure  ABC  is 
the  directrix. 

570.  Element.  The  generatrix 
in  any  position  is  called  an  ele- 
ment of  the  cylindric  surface. 

571.  Cylinder.  A  solid  bounded  by  a  cylindric  surface  and 
two  parallel  plane  surfaces  is  called  a  cylinder. 


It  follows,  therefore,  that  all  the  elements  of  a  cylinder  are  equal. 
The  terms  bases,  lateral  surface.,  and  altitude  are  used  as  with  prisms. 

572.  Right  and  Oblique  Cylinders.  A  cylinder  whose  elements 
are  perpendicular  to  its  bases  is  called  a  right  cylinder;  other- 
wise a  cylinder  is  called  an  oblique  cylinder. 

573.  Section  of  a  Cylinder.  A  figure  formed  by  the  intersec- 
tion of  a  jjlane  and  a  cylinder  is  called  a  section  of  the  cylinder. 


354  BOOK  VII.    SOLID  GEOMETRY 

PEOPOSiTioisr  XX.    Theorem 

574.  Every  section  of  a  cylinder  made  hy  a  plane 
passing  through  an  element  is  a  parallelogram. 


Given  a  cylinder  ACy  and  a  section  ABCD  made  by  a  plane  pass- 
ing through  the  element  AB. 

To  prove  that     ABCD  is  a  parallelogram. 

Proof.    Through  D  draw  a  line  in  the  plane  ABCD  II  to  AB. 
This  line  is  an  element  of  the  cylindric  surface.  §  570 

Since  this  line  is  in  both  the  plane  and  the  cylindric  surface, 
it  must  be  their  intersection  and  must  coincide  with  DC. 
Hence  DC  coincides  with  a  straight  line  parallel  to  AB. 

Therefore  i)C  is  a  straight  line  II  to  AB. 

Also  ^Z)  is  a  straight  line  II  to  BC.  §  453 

.'.  ABCD  is  a  parallelogram,  by  §  118.  q.e.d. 

575.  Corollary.    Every  section  of  a  right  cylinder  made 
hy  a  plane  passing  through  an  element  is  a  rectangle. 

576.  Circular  Cylinder.    A  cylinder  whose  bases  are  circles  is 
called  a  Gircular  cylinder. 

A  right  circular  cylinder,  being  generated  by  the  revolution  of  a  rec- 
tangle about  one  side  as  an  axis,  is  also  called  a  cylinder  of  revolution. 


CYLINDERS  355 

Proposition  XXI.    Theorem 
577.  The  bases  of  a  cylinder  are  congruent. 


A 
Given  the  cylinder  AC,  with  bases  ABE  and  DCG. 
To  prove  that       ABE  is  congruent  to  DCG. 
Proof.    Let  A,  B,  E  be  any  three  points  in  the  perimeter  of 
the  lower  base,  and  AD,  BC,  EG  be  elements  of  the  surface. 
Draw  AB,  AE,  EB,  DC,  DG,  GC. 
Then  AD,  BC,  EG  are  equal,  §  571 

and  parallel.  §  569 

.-.  AB  =  DC,  AE  =  DG,  EB  =  GC.  §  130 

.-.A  ABE  is  congruent  to  A  DCG.  §  80 

Place  the  lower  base  on  the  upper  base  so  that  the  A  ABE 
shall  fall  on  the  A  DCG.    Then  A,  B,  E  will  fall  on  D,  C,  G. 

Therefore  all  points  in  either  perimeter  will  coincide  with 
points  in  the  other,  and  the  bases  are  congruent,  by  §  66.     Q.  e.  d. 

578.  Corollary  1.  Ang  two  parallel  sections  of  a  cylinder, 
cutting  all  the  elements,  are  congruent.  ^ 

579.  Corollary  2.  Any  section  of  a  cylinder  parallel  to 
the  base  is  congruent  to  the  base. 

580.  Corollary  3.  The  straight  line  joining  the  centers  of 
the  bases  of  a  circular  cylinder  passes  through  the  centers  of 
all  sections  of  the  cylinder  parallel  to  the  bases. 


356 


BOOK  VII.    SOLID  GEOMETRY 


581.  Tangent  Plane.  A  plane  which  contains  an  element  of 
a  cyJinder,  but  does  not  cut  the  surface,  is  called  a  tangent  plane 
to  the  cylinder. 

582.  Construction  of  Tangent  Planes.  From  a  consideration 
of  the  nature  of  a  tangent  plane  and  of  the  construction  of  a 
cylindric  surface  it  is  evident  that : 

A  plane  passing  through  a  tangent  to  the  base  of  a  circular 
cylinder  and  the  element  drawn  throiujh  the  point  of  contact  is 
tangent  to  the  cylinder. 

If  a  plane  is  tangent  to  a  circular  cylinder,  its  intersection 
with  the  plane  of  the  base  is  tangent  to  the  base. 

583.  Inscribed  Prism.  A  prism  whose  lateral  edges  are  ele- 
ments of  a  cylinder  and  whose  bases  are  inscribed  in  the  bases 
of  the  cylinder  is  called  an  inscribed  prism. 

In  this  case  the  cylinder  is  said  to  be  circumscribed  about  the  prism. 


Inscribed  Prism 


Circumscribed  Prism 


584.  Circumscribed  Prism.  A  prism  whose  lateral  faces  are 
tangent  to  the  lateral  surface  of  a  cylinder  and  whose  bases 
are  circumscribed'  about  the  bases  of  the  cylinder  is  called  a 
circumscribed  prism. 

In  this  case  the  cylinder  is  said  to  be  inscribed  in  the  prism. 


CYLINDERS 


357 


585.  Right  Section.  A  section  of  a  cylinder  made  by  a  plane 
that  cuts  all  the  elements  and  is  perjiendicular  to  them  is  called 
a  right  section  of  the  cylinder. 

586.  Cylinder  as  a  Limit.  From  the  work  already  done  In 
connection  with  limits,  and  from  the  nature  of  the  inscribed 
and  circumscribed  prisms,  the  following  properties  of  the 
cylinder  may  now  be  assumed  without  further  proof  than 
that  given  below : 

If  a  prism  ivhose  base  is  a  regular  polygon  is  inscribed  in  or 
circumscrihed  about  a  circular  cylinder,  and  if  the  nurriber  of 
sides  of  the  prism  is  indefinitely  increased, 

1.  The  volume  of  the  cylinder  is  the  lirnit  of  the  volume  of 
the  prisvi. 

2.  The  lateral  area  of  the  cylinder  is  the  limit  of  the  lateral 
area  of  the  prism. 

3.  The  perimeter  of  a  right  section  of  the  cylinder  is  the  limit 
of  the  pjerimeter  of  a  right  section  of  the  prism. 


For  as  we  increase  the  number  of  sides  of  the  base  of  the  inscribed 
or  circumscribed  prism  whose  base  is  a  regular  polygon,  the  perimeter 
of  the  base  approaches  the  circle  as  its  limit  (§  381). 

This  brings  the  lateral  surface  of  each  prism  nearer  and  nearer  the 
lateral  surface  of  the  cylinder.  It  also  brings  the  volume  of  each  prism 
nearer  and  nearer  the  volume  of  the  cylinder.  In  the  same  way  it  brings 
the  right  section  of  each  prism  nearer  and  nearer  the  right  section  of 
the  cylinder. 


358  BOOK  VII.    SOLID  GEOMETRY 

Proposition  XXII.    Theorem 

587.  The  lateral  area  of  a  circular  cylinder  is  equal 
to  the  product  of  an  element  by  the  perimeter  of  a 
right  section  of  the  cylinder. 


Given  a  circular  cylinder  C,  /  being  the  lateral  area,  p  the  perim- 
eter of  a  right  section,  and  e  an  element. 

To  prove  that  I  =  ep. 

Proof.  Suppose  a  prism  with  base  a  regular  polygon  to  be 
inscribed  in  C,  I'  being  its  lateral  area  and  p'  being  the  perim- 
eter of  its  right  section. 

Then  I' =  ep'.  §512 

If  the  number  of  lateral  faces  of  the  prism  is  indefinitely 

increased, 

I'  approaches  Z  as  a  limit, 

and  p'  approaches^  as  a  limit,  §  586 

and  consequently  ep'  approaches  ep  as  a  limit. 

.-.  l  =  ep,  by  §  207.  q.e.d. 

588.  Corollary.  The  lateral  area  of  a  cylinder  of  revolu- 
tion is  equal  to  the  product  of  the  altitude  by  the  circum- 
ference of  the  base. 

In  the  case  of  a  right  circular  cylinder  of  altitude  a,  lateral  area  /, 
total  area  t,  and  radius  of  base  r,  we  have 

I  =  2  Trra,  and  t  =  2  Trra  +  2  irr^  =  2  wr  {a -{■  r) . 


CYLINDEES 


359 


Proposition  XXIII.    Theorem 

589.  The  volume  of  a  circular  cylinder  is  equal  to  the 
product  of  its  base  hy  its  altitude. 


Given  a  circular  cylinder  C,  b  being  the  base,  v  the  volume, 
and  a  the  altitude. 

To  prove  that  v  =  ba. 

Proof.    Suppose  a  prism  with  base  a  regular  polygon  to  be 
inscribed  in  C,  b'  being  its  base  and  v'  being  its  volume. 

Then  v'  =  b'a.  §  539 

If  the  number  of  lateral  faces  of  the  prism  is  indefinitely 
increased, 

v'  approaches  v  as  a  limit,  §  586 

b'  approaches  ^  as  a  limit,  §  381 

and  consequently  b'a  approaches  ba  as  a  limit. 


But 


v'=  b'a,  whatever  the  number  of  sides. 


§539 

Q.E.D. 


.'.v  =  ba,  by  §207. 

590.  Corollary.  The  volume  of  a  cylinder  of  revolution 
with  radius  r  and  altitude  a  is  irr^a. 

What  is  the  area  of  the  base  ?   By  what  is  this  to  be  multiplied  ? 

591.  Similar  Cylinders.  Cylinders  generated  by  the  revolu- 
tion of  similar  rectangles  about  corresponding  sides  are  called 
similar  cylinders  of  revolution. 

§§  691  and  692  may  be  omitted  without  destroying  the  sequence. 


360  BOOK  VII.    SOLID  GEOMETRY 

Proposition  XXIV.    Theorem 

592.  The  lateral  areas ^  or  the  total  areas,  of  similar 
cylinders  of  revolution  are  to  each  other  as  the  squares 
of  their  altitudes  or  as  the  squares  of  their  radii ;  and 
their  voluDies  are  to  each  other  as  the  cubes  of  their 
altitudes  or  as  the  cubes  of  their  radii. 


Given  two  similar  cylinders  of  revolution,  /  and  /'  denoting  their 
lateral  areas,  t  and  t'  their  total  areas,  v  and  v'  their  volumes, 
a  and  a'  their  altitudes,  and  r  and  r'  their  radii. 

To  prove  that  I:  I' =t :  t'  =  a" :  a'^  =  r^ :  r'% 

and  that  v  :  v'  =  a^ :  a'^  =  r^ :  r'\ 

Proof.    Since  the  generating  rectangles  are  similar,         §  591 

a       ?•       a  -\-r 
•'-a'^r'^^a^^''  §269 

Also  we  have  by  this  proportion  and  §  588, 
I        2  irra  _  ra  _  r'^  _  a^ 
V  ^  2  irr'a'  ^7^'~7^~'^'' 

But  t  ==  2  TTva  +  2  irr^  (§  588),  and  v  =  irr^a.  §  590 

t         2  7rr(a  +  r)    _   r(a  +  r)   _  r^  _  a^ 
■'•  ?  ~  2  irr\a^  +  r')  ~  r'(a'  +  r-')  ~  r''  ~  a""' 

V  __  irr'^a  _  ?*^        '^  _  ^^  _  ^'^ 


CYLINDERS  361 

EXERCISE  93 

1.  The  diameter  of  a  well  is  6  ft.  and  the  water  is  7  ft. 
deep.  How  many  gallons  of  water  are  there  in  the  well,  reck- 
oning 7^  gal.  to  the  cubic  foot  ? 

2.  When  a  body  is  placed  under  water  in  a  right  circular 
cylinder  60  centimeters  in  diameter,  the  level  of  the  water  rises 
40  centimeters.    Find  the  volum6  of  the  body. 

3.  How  many  cubic  yards  of  earth  must  be  removed  in 
constructing  a  tunnel  100  yd.  long,  the  section  being  a  semi- 
circle with  a  radius  of  18  ft.  ? 

4.  How  many  square  feet  of  sheet  iron  are  required  to 
make  a  pipe  18  in.  in  diameter  and  40  ft.  long  ? 

5.  Find  the  radius  of  a  cylindric  pail  14  in.  high  that  will 
hold  exactly  2  cu.  ft. 

6.  The  height  of  a  cylindric  vessel  that  will  hold  20  liters 
is  equal  to  the  diameter.    Find  the  altitude  and  the  radius. 

7.  If  the  total  surface  of  a  right  circular  cylinder  is  t  and 
the  radius  of  the  base  is  r,  find  the  altitude  a. 

8.  If  the  lateral  surface  of  a  right  circular  cylinder  is  I 
and  the  volume  is  v,  find  the  radius  r  and  the  altitude  a. 

9.  If  the  circumference  of  the  base  of  a  right  circular  cyl- 
inder is  c  and  the  altitude  is  (X,  find  the  volume  v. 

10.  If  the  circumference  of  the  base  of  a  right  circular 
cylinder  is  c  and  the  total  surface  is  ^,  find  the  volume  v. 

11.  If  the  volume  of  a  right  circular  cylinder  is  v  and  the 
altitude  is  a,  find  the  total  surface  i^. 

12.  If  V  is  the  volume  of  a  right  circular  cylinder  in  which 
the  altitude  equals  the  diameter,  find  the  altitude  a  and  the 
total  surface  t. 

13.  From  the  formula  ^^  =  2  7r/'(a  +  r)  (§  588)  find  the  value 
of  T.    (Omit  unless  quadratics  have  been  studied.) 


362  BOOK  VIL    SOLID  GEOMETRY 

593.  Conic  Surface.  A  surface  generated  by  a  straight  line 
which  constantly  touches  a  fixed  plane  curve  and  passes 
through  a  fixed  point   not  in  the   plane 

of  the  curve  is  called  a  conic  surface  or  :  lliilii 

a  conical  surface.  iiiill!l\  ^^ 

The  moving  line  is  called  the  generatrix,  the  i||!    '  '^^ 

fixed  curve  the  directrix,  and  the  fixed  point  the  W 

vertex.  f 

Hold  a  pencil  by  the  point  and  let  the  other 

end  swing  around  a  circle,  and  the  pencil  will  | 

generate  a  conic  surface.  ||||         ^ 

We   may    also    swing    a    blackboard   pointer  ;^ 

about  any  point  near  the  middle,  so  that  either  ^^ 

end  shall  touch  any  fixed  plane  curve,  and  thus  ^^^ 

generate  a  conic  surface.    Such  a  surface  is  rep-  .,  _^^         m 

resented  in  the  annexed  figure.  '■■■-i:iJi::|:iill:l!!i!!Jg^^^^^ 

594.  Element.  The  generatrix  in  any  position  is  called  an 
element  of  the  conic  surface. 

If  the  generatrix  is  of  indefinite  length,  the  surface  consists  of  two 
portions,  one  above  and  the  other  below  the  vertex,  which  are  called 
the  upper  nappe  and  lower  nappe  respectively.  The  two  nappes  are  shown 
in  the  above  figure. 

595.  Cone.  A  solid  bounded  by  a  conic  surface  and  a  plane 
cutting   all   the  elements  is  called  a  cone. 

The  conic  surface  is  called 
the  lateral  surface  of  the  cone, 
and  the  plane  surface  is  called 
the  base  of  the  cone. 

The  vertex  of  the  conic  sur- 
face is  called  the  vertex  of   the 

cone,   and   the   elements   of   the  /    i;!^^       I  /      ill 

conic  surface  are  called  the  ele- 
ments of  the  cone.  l^  lijijl 

The     perpendicular    distance        :. 
from  the  vertex  to  the  plane  of 
the  base  is  called  the  altitude  of 
the  cone. 


CONES 


363 


596.  Circular  Cone.  A  cone  whose  base  is  a  circle  is  called  a 
circular  cone. 

The  straight  line  joining  the  vertex  of  a  circular  cone  and  the  center 
of  the  base  is  called  the  axis  of  the  cone. 

597.  Right  and  Oblique  Cones.  A  circular  cone  whose  axis  is 
perpendicular  to  the  base  is  called  a  right  cone ;  otherwise  a 
circular  cone  is  called  an  oblique  cone. 

598.  Cone  of  Revolution.  Since  a  right 
circular  cone  may  be  generated  by  the 
revolution  of  a  right  triangle  about  one 
of  the  sides  of  the  right  angle,  it  is  called 
a  cone  of  revolution. 

In  this  case  the  hypotenuse  corresponds  to 
an  element  of  the  surface  and  is  called  the  slant  height. 

599.  Conic  Section.  A  section  formed  by  the  intersection  of  a 
plane  and  the  conic  surface  of  a  cone  of  revolution  is  called  a 
conic  section. 


Fig.  1 


Fig.  2 


Fig.  3 


Fig.  4 


Fig.  5 


In  Fig.  1  the  conic  section  is  two  intersecting  straight  lines,  and  this 
is  discussed  in  §  600.   This  is  true  for  all  kinds  of  cones. 

In  Fig.  2  the  conic  section  is  a  circle,  and  this  is  discussed  in  §  601. 

In  Fig.  3  the  conic  section  is  called  an  ellipse.,  the  form  a  circle  seems 
to  take  when  looked  at  obliquely.    The  orbit  of  a  planet  is  an  ellipse. 

In  Fig.  4  the  conic  section  is  a  parabola.,  the  path  of  a  projectile  (in  a 
vacuum).    Here  the  cutting  plane  is  parallel  to  an  element. 

In  Fig.  5  the  conic  section  is  an  hyperbola. 

The  general  study  of  conic  sections  is  not  a  part  of  elementaiy  geome- 
try, but  the  names  of  the  sections  may  profitably  be  known. 


364 


BOOK  VII.    SOLID  GEOMETRY 


Proposition  XXV.    Theorem 

600.  Every  section  of  a  cone  made  hy  a  plane  pass- 
ing through  its  vertex  is  a  triangle. 


I 


Given  a  cone,  with  AVB  a.  section  made  by  a  plane  passing 
through  the  vertex  V. 

To  prove  that  A  VB  is  a  triangle. 

Proof.  A  B  is  Si  straight  line.  §  429 

Draw  the  straight  lines  VA  and  VB. 

The  lines  VA .  and  VB  are  both  elements  of  the  surface  of 
the  given  cone.  §  594 

These  lines  lie  in  the  cutting  plane,  since  their  extremities 
are  in  the  plane.  §  422 

Hence  VA  and  VB  are  the  intersections  of  the  conic  surface 
with  the  cutting  plane. 

But  VA  and  VB  are  straight  lines.  Const. 

Therefore  the  intersections  of  the  conic  surface  and   the 
plane  are  straight  lines. 

Therefore  the  section  A  VB  is  a  triangle,  by  §  28.  q.e.d. 


CONES 


365 


Proposition  XXVI.   Theorem 

601.  In  a  circular  cone  a  section  made  hy  a  plane 
parallel  to  the  base  is  a  circle. 


Given  the  circular  cone  V-ABCD^  with  the  section  A'B'C'D' 
parallel  to  the  base. 

To  prove  that         A'B'C'I)'  is  a  circle. 

Proof.    Let  0  be  the  center  of  tlie  base,  and  let  0'  be  the  point 

in  which  the  axis  VO  pierces  the  plane  of  the  conic  section. 
Through  VO  and  any  elements  VA,  VB,  pass  planes  cutting 

the  base  in  the  radii  OA,  OB,  and  cutting  the  section  A'B'C'D' 

in  the  straight  lines  O'A',  O'B'. 

Then  O'yl'  and  O'B'  are  II  respectively  to  OA  and  OB.        §  453 
Therefore  the  A  ^  0  F  and  OB  V  are  similar  respectively  to 

the  AA'O'V  and  O'B'V.  §  285 

,OA_^VO_^OB_ 
"O'A'       VO'       O'B''  ^ 

But  0A  =  OB.  §  162 

.-.  0'.4'=0'i5'(§263),and^'5'C'Z>'isacircle,by  §159.  q.e.d. 

602.  Corollary.    The  axis  of  a  circular  cone  passes  through 
the  center  of  every  section  which  is  parallel  to  the  base. 


366 


BOOK  VIL    SOLID  GEOMETRY 


603.  Tangent  Plane.  A  plane  which  contains  an  element  of 
a  cone,  but  does  not  cut  the  surface,  is  called  a  tangent  plane 
to  the  cone. 

604.  Construction  of  Tangent  Planes. 

It  is  evident  that : 

A  plane  passing  through  a  tangent  to 
the  base  of  a  circular  cone  and  the  ele- 
ment drawn  through  the  point  of  contact 
is  tangent  to  the  cone. 

If  a  plane  is  tangent  to  a  circular 
cone  its  intersection  with  the  plane  of 
the  base  is  tangent  to  the  base. 

605.  Inscribed  Pyramid.  A  pyramid  whose  lateral  edges  are 
elements  of  a  cone  and  whose  base  is  inscribed  in  the  base  of 
the  cone  is  called  an  inscribed  pyrainid. 

In  this  case  the  cone  is  said  to  be  circumscribed  about  the  pyramid. 


Inscribed  Pyramid  Circumscribed  Pyramid 

606.  Circumscribed  Pyramid.  A  pyramid  whose  lateral  faces 
are  tangent  to  the  lateral  surface  of  a  cone  and  whose  base 
is  circumscribed  about  the  base  of  the  cone  is  called  a  cir- 
cumscribed pyramid. 

In  this  case  the  cone  is  said  to  be  inscribed  in  the  pyramid. 


CONES 


367 


607.  Frustum  of  a  Cone.  The  portion  of  a  cone  included  be- 
tween the  base  and  a  section  parallel  to  the  base  is  called  a 
friistuvi  of  a  cone. 

The  base  of  the  cone  and  the  parallel  section 
are  together  called  the  hases  of  the  frustum. 

The  terms  altitude  and  lateral  area  of  a  f  rus-     _ 
turn  of  a  cone,  and  slant  height  of  a  frustum 
of  a  right  circular  cone,  are  used  in  substan- 
tially the  same  manner  as  with  the  frustum  of 
a  pyramid  (§§  550,  551,  552). 

608.  Cones  and  Frustums  as  Limits.  The  following  proper- 
ties, similar  to  those  of  §  586,  are  assumed  without  proof : 

If  a  pyramid  whose  base  is  a  regular  polygon  is  inscribed  in 
or  circurnscrihed  about  a  circular  cone,  and  if  the  number  of 
sides  of  the  base  of  the  pyramid  is  indefinitely  increased,  the 
volume  of  the  cone  is  the  limit  of  the  volume  of  the  pyrarnid, 
and  the  lateral  area  of  the  cone  is  the  limit  of  the  lateral  area 
of  the  pyramid. 


The  volume  of  a  frustum  of  a  cone  is  the  limit  of  the  volumes 
of  the  frustums  of  the  inscribed  and  circu7nscribed  pyramids, 
if  the  number  of  lateral  faces  is  indefinitely  increased,  and 
the  lateral  area  of  the  frustum  of  a  cone  is  the  lim.it  of  the 
lateral  areas  of  the  frustums  of  the  inscribed  and  circumscribed 
pyramids,  the  bases  being  regular  polygons. 


368  BOOK  VII.    SOLID  GEOMETRY 

Proposition  XXVII.    Theorem 

609.  The  lateral  area  of  a  cone  of  i^evolution  is  equal 
to  half  the  product  of  the  slant  height  by  the  circumfer- 
ence of  the  base. 


Given  a  cone  of  lateral  area  /,  circumference  of  base  c,  and  slant 
height  s. 

To  prove  that  1=  '^  so. 

Proof.    Suppose  a  regular  pyramid  to  be  circumscribed  about 

the  cone,  the  perimeter  of  its  base  being  2^  and  its  lateral  area  I'. 

Then  l'  =  ^sp.  §553 

If  the  number  of  the  lateral  faces  of  the  circumscribed  pyra- 
mid is  indefinitely  increased, 

V  approaches  Z  as  ^  limit,  §  608 

p  approaches  c  as  a  limit,             '  §  381 
and  consequently  ^  sp  approaches  1  sc  as  a  limit. 

But              ^'  =  2  ^^P?  whatever  the  number  of  sides.  §  553 

.-.  Z  =^sc,  by  §  207.  Q.e.d. 

610.  Corollary.  If  I  denotes  the  lateral  area,  t  the  total 
area,  s  the  slant  height,  and  r  the  radius  of  the  base  of  a  cone 
of  revolution,  then 

Z  =  1(2  7rr  X  s)  =  irrs  ; 

t  =  irrs  +  Trr^  =  7rr  (s  +  r). 


CONES  369 

EXERCISE  94 

Find  the  lateral  areas  of  cones  of  revolution,  given  the  slant 
heights  and  the  circumferences  of  the  bases  respectively  as 
follows : 

1.  21  in.,  ^  in.       4.  3.7  in.,  5.8  in.       7.  2  ft.  6  in.,  4  ft.  8  in. 

2.  4|  in.,  8i  in.       5.  5.3  in.,  9.7  in.      8.  3  ft.  7  in.,  8  ft.  6  in. 

3.  6y5^  in.,  10^  in.  6.  6.5  in.,  11.6  in.    9.  5  ft.  8  in.,  12  ft.  4  in. 

Find  the  lateral  areas  of  cones  of  revolution,  given  the  slant 
heights  and  the  radii  of  the  bases  respectively  as  follows : 

10.  3|  in.,  2\  in.     13.  6.4  in.,  4.8  in.     16.  2  ft.  3  in.,  8  in. 

11.  2^  in..  If  in.     14.  7.2  in.,  5.3  in.     17.  4  ft.  6  in.,  2  ft. 

12.  ^  in.,  31  in.     15.  8.9  in.,  b.Q  in.     18.  6  ft.  9  in.,  3  ft.  2  in. 

Find  the  total  areas  of  cones  of  revolution,  given  the  slant 
heights  and  the  radii  of  the  bases  respectively  as  follows : 

19.  3  in.,  2  in.         21.  7  in.,  4  in.  23.  6  ft.,  4  ft. 

20.  5  in.,  3  in.         22.  9  in.,  5  in.  24.  12  ft.,  5  ft. 

25.  Deduce  a  formula  for  finding  the  lateral  area  of  a  cone  of 
revolution  in  terms  of  the  radius  of  the  base  and  the  altitude. 

26.  Deduce  a  formula  for  finding  the  slant  height  in  terms 
of  the  lateral  area  and  the  circumference  of  the  base. 

27.  Deduce  a  formula  for  finding  the  slant  height  in  terms 
of  the  lateral  area  and  the  radius  of  the  base. 

28.  Deduce  a  formula  for  finding  the  radius  of  the  base  in 
terms  of  the  lateral  area  and  the  slant  height. 

29.  Deduce  a  formula  for  finding  the  slant  height  in  terms 
of  the  total  area  and  the  radius  of  the  base. 

30.  Deduce  a  formula  for  finding  the  circumference  of  the 
base  in  terms  of  the  lateral  area  and  the  slant  height. 


370  BOOK  VIL    SOLID  GEOMETEY 

Proposition  XXVIII.    Theorem 

611.  The  volume  of  a  circular  cone  is  equal  to  one 
third  the  product  of  its  base  by  its  altitude. 


Given  a  circular  cone  of  volume  i;,  base  &,  and  altitude  a. 

To  prove  that  v  =  ^ba. 

Proof.  Suppose  a  pyramid  with  base  a  regular  polygon  to  be 
inscribed  in  the  cone,  b'  being  its  base  and  v'  its  volume. 

Then  v'  =  ^h'a.  §561 

If  the  number  of  lateral  faces  of  the  pyramid  is  indefinitely 

increased, 

v'  approaches  v  as  a  limit,  §  608 

b'  approaches  Z>  as  a  limit,  §  381 

and  consequently  b'a  approaches  ba  as  a  limit. 

.-.  V  =  :^  ba,  hy  ^  207.  Q.e.d. 

612.  Corollary.  In  a  circular  cone  of  radius  r  and  alti- 
tude a,  V  =  ^  irr^a. 

For  the  area  of  the  base  is  irr^  (§  389). 

613.  Similar  Cones.  Cones  generated  by  the  revolution  of 
similar  right  triangles  about  corresponding  sides  are  called 
similar  cones  of  revolution. 

In  case  §  614  is  omitted  this  definition  may  also  be  omitted. 


CONES  371 

EXERCISE  95 

Find  the  volumes  of  circular  cones,  given  the  altitudes  and 
the  areas  of  the  bases  respectively/  as  follows  : 

1.  4  in.,  8  sq.  in.  4.  6.3  in.,  3.8  sq.  in. 

2.  Si  in.,  9f  sq.  in.  5.  7.8  in.,  6.9  sq.  in. 

3.  5|  in.,  10^  sq.  in.  6.  9.3  in.,  16.8  sq.  in. 

Find  the  volumes  of  circular  cones,  given  the  altitudes  and 
the  radii  of  the  bases  respectively/  as  follows : 

7.  4  in.,  3  in.  10.  9.8  in.,  4.3  in. 

8.  6  in.,  4  in.  11.  10.5  in.,  6.2  in. 

9.  8  in.,  5  in.  12.  14.9  in.,  9.6  in. 

13.  How  many  cubic  feet  in  a  conical  tent  10  ft.  in  diameter 
■and  7  ft.  high? 

14.  How  many  cubic  feet  in  a  conical  pile  of  earth  15  ft.  in 
diameter  and  8  ft.  high  ? 

15.  Deduce  a  formula  for  finding  the  altitude  of  a  circular 
cone  in  terms  of  the  volume  and  the  area  of  the  base. 

16.  Deduce  a  formula  for  finding  the  area  of  the  base  of  a 
circular  cone  in  terms  of  the  volume  and  the  altitude. 

17.  Deduce  a  formula  for  finding  the  altitude  of  a  circular 
cone  in  terms  of  the  volume  and  the  radius  of  the  base. 

18.  Deduce  a  formula  for  finding  the  radius  of  the  base  of 
a  circular  cone  in  terms  of  the  volume  and  the  altitude. 

19.  Deduce  a  formula  for  finding  the  volume  of  a  cone  of 
revolution  in  terms  of  the  slant  height  and  the  radius  of  the 
base. 

20.  Deduce  formulas  for  finding  the  slant  height  and  the 
altitude  of  a  cone  of  revolution  in  terms  of  the  volume  and  the 
radius  of  the  base. 


372  BOOK  VII.    SOLID  GEOMETRY 

Proposition  XXIX.    Theorem 

614.  The  lateral  areas,  or  the  total  areas,  of  tivo  sim- 
ilar cones  of  revolution  are  to  each  other  as  the  squares 
of  their  altitudes,  as  the  squares  of  their  radii,  or  as  the 
squares  of  their  slant  heights  ;  and  their  volumes  are 
to  each  other  as  the  cubes  of  their  altitudes,  as  the  cubes 
of  their  radii,  or  as  the  cubes  of  their  slant  heights. 


Given  two  similar  cones  of  revolution,  with  lateral  areas  /  and 
/',  total  areas  t  and  t\  volumes  v  and  v\  altitudes  a  and  a',  radii 
r  and  r\  and  slant  heights  5  and  s'  respectively. 


To  prove 

that  1:1'  = 

t:t'  =  a':a 

'■^  =  r' 

'.,^n 

'  =  s' 

:s'% 

id  that 

V  :  v'  = 

a':a"  =  r': 

/«  = 

s':. 

s". 

Proof. 

a 
a' 

r        s 

s  +  r 

§§282,269 

I 
V~ 

irrs  _r 
irr's'  ~  r'  "" 

§610 

t 

irrCs  +  r) 
irrXs'-j-r') 

,,2 

s' 
s'-' 

a' 

a'' 

• 

§610 

V 

^  Trr'ht'  "  r" 

a        7'^ 

a' 

7^' 

by  § 

612. 

Q.E.D. 

§§  613  and  614,  like  §§  591  and  592,  are  occasionally  demanded  in 
college  entrance  examinations.  They  are  not  needed  for  any  exercises 
and  they  may  be  omitted  without  destroying  the  sequence. 


CONES  373 

Proposition  XXX.    Theorem 

615.  The  lateral  area  of  a  frustum  of  a  cone  of  revo- 
lution is  equal  to  half  the  sum  of  the  circumferences  of 
its  bases  multiplied  by  the  slant  height. 


Given  a  frustum  of  a  cone  of  revolution,  with  lateral  area  /, 
circumferences  of  bases  c  and  c',  and  slant  height  s. 

To  prove  that  Z  =  |-  («?  -f-  c'^s. 

Proof.  Suppose  a  frustum  of  a  regular  pyramid  circum- 
scribed about  the  frustum  of  the  cone,  as  a  pyramid  is  cir- 
cumscribed about  a  cone. 

Let  the  lateral  area  of  the  circumscribed  frustum  be  I',  and 
let  p  and  p'  be  the  perimeters  of  the  bases  corresponding  to 
c  and  c'  respectively.  The  slant  height  is  s,  the  same  as  that 
of  the  frustum  of  the  cone. 

Then  l'  =  l(p^p')s.  §554 

If  the  number  of  lateral  faces  of  the  circumscribed  frustum 
is  indefinitely  increased,  what  limits  do  I'  and^  +^>'  approach  ? 
Therefore  what  limit  does  ^(p  -{-p')s  approach  ? 
What  conclusion  may  be  drawn,  as  in  §  587  ? 
Complete  the  proof. 

616.  Corollary.  The  lateral  area  of  a  frustum  of  a  cone 
of  revolution  is  equal  to  the  circumference  of  a  section  equi- 
distant from  its  bases  multiplied  hy  its  slant  heiyht. 

How  can  it  be  proved  that  \{c  +  c')  equals  the  circumference  of  this 
section  ?    How  are  the  radii  related  ? 


374  BOOK  VII.    SOLID  GEOMETRY 

Proposition  XXXI.    Theorem 

617.  A  frustum  of  a  circular  cone  is  equivalent  to 
the  sum  of  three  cones  whose  common  altitude  is  the 
altitude  of  the  frustum  and  whose  bases  are  the  lower 
base,  the  upper  base,  and  the  mean  p)Toportional  between 
the  bases  of  the  frustum. 


Given  a  frustum  of  a  circular  cone,  with  volume  v,  bases  &and  6', 
and  altitude  a. 

To  prove  that       v  =  \a{h-{-y  +  Vw). 

Proof.  Suppose  a  frustum  of  a  pyramid  with  base  a  regular 
polygon  to  be  inscribed  in  the  frustum  of  the  cone,  as  a  pyramid 
is  inscribed  in  a  cone. 

Let  v'  be  the  volume,  and  let  x  and  a?'  be  the  bases  corre- 
sponding to  h  and  Z>'  respectively.  The  altitude  is  a,  the  same 
as  that  of  the  frustum  of  the  cone. 

Then  v' =  la{x-\-x' -{- -Vxx').  §565 

If  the  number  of  lateral  faces  of  the  inscribed  frustum  is  in- 
definitely increased,  what  limits  do  v\  x,  x',  and  xx'  approach  ? 
Therefore  what  limit  does  ^  a{x  -\-  x'  -\-  -Vxx')  approach  ? 
What  conclusion  may  be  drawn  ? 
Complete  the  proof. 

618.  Corollary.   In   a  frustum   of  a   cone  of  revolution^ 
r  and  r^  being  the  radii  of  the  hase.%  v  =  ^  ira^f'  -f-  r'^  +  rr^). 
For  6  =  7rr2,  h'  —  irr'".     .-.  Vhb'  =  •VTrr^XTrr^  =  Trrr". 


CONES  375 

EXERCISE  96 

Find  the  lateral  areas  of  frustums  of  coneSj  given  the  cir- 
cumferences of  the  bases  and  the  slant  heights  respectively  as 
follows  : 

1.  c  =  4  in.,  c'  =  3  in.,  s  =  0.5  in. 

2.  c  =  6  in.,  g'  =  5  in.,  s  =  1.4  in. 

3.  c  =  7^  in.,  c'  =z  5f  in.,  s  =  2^  in. 

4.  c  =  23  in.,  c'  =  18  in.,  s  =  16  in. 

Find  to  two  decimal  places  the  volumes  of  frustums  of  cones, 
given  the  altitudes  and  the  areas  of  the  bases  respectively  as 
follows : 

5.  a  =  S  in.,  b  =  A^  sq.  in.,  b'  =  2  sq.  in. 

6.  a  =  4  in.,  ^  —  8^  sq.  in.,  b'  =  3  sq.  in. 

7.  a  =  5^  in.,  ^  =  16  sq.  in.,  ^'  =  9sq.  in. 

8.  a  =  6  in.,  b  =  17  sq.  in.,  b'  =  11  sq.  in. 

Find  to  two  decimal  places  the  volumes  of  frustums  of  cones 
of  revolution,  given  the  altitudes  and  the  radii  of  the  bases 
respectively/  as  follows  : 

9.  a  =  4:  in.,  r  =  3  in.,  r'  =  2  in. 

10.  a  =  5  in.,  r  =  3^  in.,  ?•'  =  2i  in. 

11.  a=6  in.,  r  =  3.7  in.,  r'  =  3.1  in. 

12.  a  =  7^  in.,  r  =  4|  in.,  r'  =  3^  in. 

13.  Deduce  a  formula  for  finding  the  altitude  of  a  frustum 
of  a  circular  cone  in  terms  of  the  volume  and  the  areas  of  the 
bases. 

14.  Deduce  a  formula  for  finding  the  altitude  of  a  frustum  of 
a  cone  of  revolution  in  terms  of  the  volume  and  the  radii  of  the 
bases. 


376  BOOK  VII.    SOLID  GEOMETRY 

EXERCISE  97 
Industkial  Pkoblems 

1.  There  is  a  rule  for  calculating  the  strongest  beam  that 
can  be  cut  from  a  cylindric  log,  as  follows : 

Erect  perpendiculars  MD  and  NB  on  opposite 
sides  of  a  diameter  A  C,  at  the  trisection  points  AI 
and  N,  meeting  the  circle  in  D  and  B.  Then 
ABCD  is  a  section  of  the  beam. 

Calculate  the  dimensions,  the  log  being  16  in.  in  diameter. 

2.  A  cylindric  funnel  for  a  steamboat  is  4  ft.  3  in.  in  diam- 
eter. It  is  built  up  of  four  plates  in  girth,  and  the  lap  of  each 
joint  is  1|  in.    Find  one  dimension  of  each  plate. 

3.  A  tubular  boiler  has  124  tubes  each  3|  in.  in  diameter 
and  18  ft.  long.  Eequired  the  total  tube  surface.  Answer  to 
the  nearest  square  foot. 

4.  A  room  in  a  factory  is  heated  by  steam  pipes.  There  are 
235  ft.  of  2-inch  pipe  and  26  ft.  3  in.  of  3-inch  pipe,  besides  2  ft. 
8  in.  of  4 1 -inch  feed  pipe.  Required  the  total  heating  surface. 
Answer  to  the  nearest  square  foot. 

5.  A  triangular  plate  of  wrought  iron  |  in.  thick  is  2  ft.  7  in. 
on  each  side.  If  the  weight  of  a  plate  1  ft.  square  and  \  in. 
thick  is  5  lb.,  find  to  the  nearest  pound  the  weight  of  the  given 
triangular  plate. 

6.  The  water  surface  of  an  upright  cylindric  boiler  is  2  ft. 
8  in.  below  the  top  of  the  boiler,  and  is  12.57  sq.  ft.  in  area. 
What  is  the  volume  of  the  steam  space  ? 

7.  A  cylinder  16  in.  in  diameter  is  required  to  hold  50  gal. 
of  water.  What  must  be  its  height,  to  the  nearest  tenth  of  an 
inch,  allowing  231  cu.  in.  to  the  gallon  ? 

8.  How  many  square  feet  of  tin  are  required  to  make  a 
funnel,  if  the  diameters  of  the  top  and  bottom  are  30  in.  and 
15  in.  respectively,  and  the  height  is  25  in.  ? 


EXERCISES 


377 


9.  Find  to  two  decimal  places  the  weight  of  a  steel  plate 
4  ft.  by  3  ft.  2  in.  by  If  in.,  allowing  490  lb.  per  cubic  foot. 

10.  A  steel  plate  for  a  steamship  is  5  ft.  long,  3  ft.  6  in. 
wide,  and  ^  in.  thick.  A  porthole  10  in.  in  diameter  is  cut 
through  the  plate.  Required  the  weight  of  the  finished  plate, 
allowing  0.29  lb.  per  cubic  inch.    Answer  to  two  decimal  places. 

11.  A  cast-iron  base  for  a  column  is  in  the  form  of  a  frus- 
tum of  a  pyramid,  the  lower  base  being  a  square  2  ft,  on  a  side, 
and  the  upper  base  having  a  fourth  of  the  area  of  the  lower 
base.  The  altitude  of  the  frustum  is  9  -in.  Required  the  weight 
to  the  nearest  pound,  allowing  460  lb.  per  cubic  foot. 

12.  A  cylinder  head  for  a  steam 
engine  has  the  shape  shown  in  the 
figure,  where  the  dimensions  in 
inches  are  :  a  =  12,  b  =  3,  c  =  2, 
d  =  6,  e  =  3,f=^,  (/  =  I,  and  A  — |. 


There  are  six  |-inch  holes  for  bolts. 


I  holes 


Compute  the  weight  of  the  plate, 

allowing  41  lb.  for  the  weight  of  a 

steel  plate  1  ft.  square  and  1  in.  thick.   Answer  to  the  nearest 

tenth  of  a  pound. 

13.  A  steel  beam  10  in.  by  5  in.,  in  the  form  here  shown,  is 
18  ft.  long.    The  thickness  of  the  beam  is  |  in.  and 

the  average  thickness  of  the  flanges  is  |  in.    Find    -^-t^ 
the  weight  of  the  beam  to  the  nearest  pound,  allow-    4    >^-5." 
ing  0.29  lb.  per  cubic  inch.  i  „ 

14.  A  hollow  steel  shaft  12  ft.  long  is  18  in.  in 
exterior  diameter  and  8  in.   in  interior  diameter.    Find   the 
weight  to  the  nearest  pound,  allowing  0.29  lb.  per  cubic  inch. 

15.  Find  the  expense,  at  70  cents  a  square  foot,  of  polishing 
the  curved  surface  of  a  marble  column  in  the  shape  of  the  frus- 
tum of  a  right  circular  cone  whose  slant  height  is  12  ft.  and  the 
radii  of  whose  bases  are  3  ft.  6  in.  and  2  ft.  4  in.  respectively. 


378  BOOK  VII.    SOLID  GEOMETRY 

EXERCISE  98 
Miscellaneous  Problems 

1.  The  slant  height  of  the  frustum  of  a  regular  pyramid  is 
25  ft.,  and  the  sides  of  its  square  bases  are  54  ft.  and  24  ft. 
respectively.    Find  the  volume. 

2.  If  the  bases  of  the  frustum  of  a  pyramid  are  regular 
hexagons  whose  sides  are  1  ft.  and  2  ft.  respectively,  and  the 
volume  of  the  frustum  is  12  cu.  ft.,  find  the  altitude. 

3.  From  a  right  circular  cone  whose  slant  height  is  30  ft., 
and  the  circumference  of  whose  base  is  10  ft.,  there  is  cut  off 
by  a  plane  parallel  to  the  base  a  cone  whose  slant  height  is 
6  ft.    Find  the  lateral  area  and  the  volume  of  the  frustum. 

4.  Find  the  difference  between  the  volume  of  the  frustum 
of  a  pyramid  whose  altitude  is  9  ft.  and  whose  bases  are 
squares,  8  ft.  and  6  ft.  respectively  on  a  side,  and  the  volume 
of  a  prism  of  the  same  altitude  whose  base  is  a  section  of  the 
frustum  parallel  to  its  bases  and  equidistant  from  them. 

5.  A  Dutch  stone  windmill  in  the  shape  of  the  frustum  of  a 
right  cone  is  12  meters  high.  The  outer  diameters  at  the  bottom 
and  the  top  are  16  meters  and  12  meters,  the  inner  diameters 
12  meters  and  10  meters.  How  many  cubic  meters  of  stone 
were  required  to  build  it  ? 

6.  The  chimney  of  a  factory  has  the  shape  of  a  frustum  of  a 
regular  pyramid.  Its  height  is  180  ft.,  and  its  upper  and  lower 
bases  are  squares  whose  sides  are  10  ft.  and  16  ft.  respectively. 
The  flue  throughout  is  a  square  whose  side  is  7  ft.  How  many 
cubic  feet  of  material  does  the  chimney  contain  ? 

7.  Two  right  triangles  with  bases  15  in.  and  21  in.,  and 
with  hypotenuses  25  in.  and  35  in.  respectively,  revolve  about 
their  third  sides.  Find  the  ratio  of  the  total  areas  of  the  solids 
generated  and  find  their  volumes. 


EXERCISES  379 

EXERCISE  99 
Equivalent  Solids 

1.  A  cube  each  edge  of  which  is  12  in.  is  transformed  into 
a  right  prism  whose  base  is  a  rectangle  16  in.  long  and  12  in. 
wide.  Find  the  height  of  the  prism  and  the  difference  between 
its  total  area  and  the  total  area  of  the  cube. 

2.  The  dimensions  of  a  rectangular  parallelepiped  are  a,  h,  c. 
Find  the  height  of  an  equivalent  right  circular  cylinder, 
having  a  for  the  radius  of  its  base  ;  the  height  of  an  equivalent 
right  circular  cone  having  a  for  the  radius  of  its  base. 

3.  A  regular  pyramid  12  ft.  high  is  transformed  into  a  regu- 
lar prism  with  an  equivalent  base.    Find  the  height  of  the  prism. 

4.  The  diameter  of  a  cylinder  is  14  ft.  and  its  height  8  ft. 
Find  the  height  of  an  equivalent  right  prism,  the  base  of  which 
is  a  square  with  a  side  4  ft.  long. 

5.  If  one  edge  of  a  cube  is  e,  what  is  the  height  h  of  an 
equivalent  right  circular  cylinder  whose  radius  is  r  ? 

6.  The  heights  of  two  equivalent  right  circular  cylinders 
are  in  the  ratio  4:9.  If  the  diameter  of  the  first  is  6  ft., 
what  is  the  diameter  of  the  second  ? 

7.  A  right  circular  cylinder  6  ft.  in  diameter  is  equivalent 
to  a  right  circular  cone  7  ft.  in  diameter.  If  the  height  of  the 
cone  is  8  ft.,  what  is  the  height  of  the  cylinder  ? 

8.  The  frustum  of  a  regular  pyramid  6  ft.  high  has  for  bases 
squares  5  ft.  and  8  ft.  on  a  side.  Find  the  height  of  an  equiva- 
lent regular  pyramid  whose  base  is  a  square  12  ft.  on  a  side. 

9.  The  frustum  of  a  cone  of  revolution  is  5  ft.  high  and 
the  diameters  of  its  bases  are  2  ft.  and  3  ft.  respectively.  Find 
the  height  of  an  equivalent  right  circular  cylinder  whose  base 
is  equal  in  area  to  the  section  of  the  frustum  made  by  a  plane 
parallel  to  the  bases  and  equidistant  from  them. 


880  BOOK  VIL    SOLID  GEOMETRY 

EXERCISE  100 
Review  Questions 

1.  Define  polyhedron.    Is  a  cylinder  a  polyhedron  ? 

2.  Define  prism,  and  classify  prisms  according  to  their  bases. 

3.  How  is  the  lateral  area  of  a  prism  computed  ?  Is  the 
method  the  same  for  right  as  for  oblique  prisms  ? 

4.  Define  parallelepiped ;  rectangular  parallelepiped ;  cube. 
Is  a  rectangular  parallelepiped  always  a  cube  ?  Is  a  cube 
always  a  rectangular  parallelepiped  ? 

5.  Distinguish  between  equivalent  and  congruent  solids. 
Are  two  cubes  with  the  same  altitudes  always  equivalent  ? 
always  congruent?    Is  this  true  for  parallelepipeds? 

6.  What  are  the  conditions  of  congruence  of  two  prisms  ? 
of  two  right  prisms  ?  of  two  cubes  ? 

7.  The  opposite  angles  of  a  parallelogram  are  equal.  What 
is  a  corresponding  proposition  concerning  parallelepipeds  ? 

8.  How  do  you  find  the  volume  of  a  parallelepiped  ?  What 
is  the  corresponding  proposition  in  plane  geometry  ? 

9.  How  do  you  find  the  volume  of  a  prism  ?  of  a  cylinder  ? 
of  a  pyramid  ?   of  a  cone  ? 

10.  Define  pyramid.  How  many  bases  has  a  pyramid  ?  Is 
there  any  kind  of  a  pyramid  in  which  more  than  one  face 
may  be  taken  as  the  base  ? 

11.  How  do  you  find  the  lateral  area  of  a  pyramid  ?  of  a  right 
cone  ?  of  a  frustum  of  a  pyramid  ?  of  a  frustum  of  a  right  cone  ? 

12.  How  many  regular  convex  polyhedrons  are  possible  ? 
What  are  their  names  ? 

13.  Given  the  radius  of  the  base  and  the  altitude  of  a  cone 
of  revolution,  how  do  you  find  the  volume  ?  the  lateral  area  ? 
the  total  area  ? 


BOOK  VIII 

THE   SPHERE 

619.  Sphere.  A  solid  bounded  by  a  surface  all  points  of 
which  are  equidistant  from  a  point  within  is  called  a  sphere. 

.  The  point  within,  from  which  all  points  on  the  surface  are  equally 
distant,  is  called  the  center.  The  surface  is  called  the  spherical  surface^ 
and  sometimes  the  sphere.  Half  of  a  sphere  is  called  a  hemisphere.  The 
terms  radius  and  diameter  are  used  as  in  the  case  of  a  circle. 

620.  Generation  of  a  Spherical  Surface.  By  the  definition  of 
sphere  it  appears  that  a  spherical  surface  may  be  generated  by 
the  revolution  of  a  semicircle  about  its  diameter  as  an  axis. 

Thus,  if  the  semicircle  ACB  revolves  about  AB,  a  spherical  surface  is 
generated.  It  is  therefore  assumed  that  a  sphere  may  be  described  with 
any  given  point  as  a  center  and  any  given  line  as  a  radius. 


621.  Equality  of  Radii  and  Diameters.    It  follows  that : 

All  radii  of  the  same  sphere  are  equal,  and  all  diameters  of 

the  same  sphere  are  equal. 

Equal  spheres  have  equal  radii,  and  spheres  having  equal 

radii  are  equal. 

381 


382 


BOOK  VIII.    SOLID  GEOMETRY 


Proposition  I.    Theorem 

622.  Every   intersection  of  a  spherical  surface  hy  a 
plane  is  a  circle. 


Given  a  sphere  with  center  0,  and  ABD  any  section  of  its 
surface  made  by  a  plane. 

To  prove  that  the  section  ABD  is  a  circle. 

Proof.  Draw  the  radii  OA,  OB,  to  any  two  points  A,  B,  in 
the  section,  and  draw  OC  A.  to  the  plane  of  the  section. 

Then  in  A  OCA  and  OCB,  A  OCA  and  OCB  are  rt.  Zs,  §  430 

OC  is  common,  and  OA  =  OB.  §  621 

.-.A  OCA  is  congruent  to  A  OCB.  §  89 

.-.  CA  =  CB.  §  67 

.  • .  any  points  A  and  B,  and  hence  all  points,  in  the  section  are 
equidistant  from  C,  and  ABD  is  a  O,  by  §159.  q.e.d. 

623.  Corollary  1.  The  line  through  the  center  of  a  sphere 
and  the  center  of  a  circle  of  the  sphere  is  perpendicular  to  the 
plane  of  the  circle. 

624.  Corollary  2.  Circles  of  a  sphere  made  hy  planes 
equidistant  from  the  center  are  equal;  and  of  two  circles  made 
hy  planes  not  equidistant  from  the  center  the  one  made  hy  the 
plane  nearer  the  center  is  the  greater. 


PLANE  SECTIONS  383 

625.  Great  Circle.  The  intersection  of  a  spherical  surface  by 
a  plane  jmssing  through  the  center  is  called  a  great  circle  of 
the  sphere. 

626.  Small  Circle.  The  intersection  of  a  spherical  surface  by 
a  plane  which  does  not  pass  through  the  center  is  called  a 
small  circle  of  the  sphere. 

627.  Poles  of  a  Circle.  If  a  diameter  of  a  sphere  is  perpen- 
dicular to  the  plane  of  a  circle  of  the  sphere,  the  extremities 
are  called  the  poles  of  the  circle. 

628.  Corollary  1.   Parallel  circles  have  the  same  poles. 

629.  Corollary  2.    All  great  circles  of  a  sphere  are  equal. 

630.  Corollary  3.  Every  great  circle  bisects  the  spherical 
surface. 

631.  Corollary  4.    Two  great  circles  bisect  each  other. 
The  intersection  of  the  planes  passes  through  what  point  ? 

632.  Corollary  5.  If  the  planes  of  two  great  circles  are 
perpendicular^  each  circle  passes  through  the  poles  of  the  other. 

Draw  the  figure  and  state  the  reason. 

633.  Corollary  6.  Through  two  given  points  on  the  sur- 
face of  a  sphere  an  arc  of  a  great  circle  may  always  be  drawn. 

Do  these  two  points,  together  with  the  center  of  the  sphere,  generally 
determine  a  plane  ?  Consider  the  special  case  in  which  the  two  points 
are  ends  of  a  diameter. 

634.  Corollary  7.  Through  three  given  points  on  the  sur- 
face of  a  sphere  one  circle  and  only  one  can  be  drawn. 

How  many  points  determine  a  plane  ? 

635.  Spherical  Distance.  The  length  of  the  smaller  arc  of 
the  great  circle  joining  two  points  on  the  surface  of  a  sphere 
is  called  the  spherical  distance  between  the  points,  or,  where 
no  confusion  is  likely  to  arise,  simply  the  distance. 


384:  BOOK  VIIL    SOLID  GEOMETEY 

Proposition  II.    Theokem 

636.  The  sjjJierical  distances  of  all  points  on  a  circle 
of  a  sioliere  from  either  pole  of  the  circle  are  equal. 


< 

/r-^-^ 

B 

1    . ■ — 

Uy 

Given  P,  P',  the  poles  of  the  circle  ABC^  and  ^,  P,  C,  any  points 
on  the  circle. 

To  'prove  that  the  great-circle  arcs  PA,  PB,  PC  are  equal. 

Proof.    The  straight  lines  PA,  PB,  PC  are  equal  §  439 

Therefore  the  arcs  PA,  PB,  PC  are  equal,  by  §  172.       q.e.d. 

In  like  manner,  the  great-circle  arcs  P'A,  P'B,  P'C  may  be  proved 
equal. 

637.  Polar  Distance.  The  spherical  distance  from  the  nearer 
pole  of  a  circle  to  any  point  on  the  circle  is  called  the  2^olar 
distance  of  the  circle. 

The  spherical  distance  of  a  great  circle  from  either  of  its  poles  may 
be  taken  as  the  polar  distance  of  the  circle. 

638.  Quadrant.  One  fourth  of  a  great  circle  is  called  a 
q}uidrant. 

639.  Corollary  1.  The  polar  distance  of  a  great  circle  is 
a  (fiadrant. 

640.  Corollary  2.  The  straight  lines  joining  points  on  a 
circle  to  either  pole  of  the  circle  are  equal. 


PLANE  SECTIONS  AND  TANGENT  PLANES    385 

Proposition  III.    Theorem 

641.  A  point  on  a  sphere,  which  is  at  the  distance  of 
a  quadrant  from  each  of  two  other  points,  not  the  ex- 
tremities of  a  diam^eter,  is  a  pole  of  the  great  circle 
passing  through  these  p)oints. 


Given  a  point  P  on  a  sphere,  PA  and  PB  quadrants,  and  ABC  the 
great  circle  passing  through  A  and  B. 

To  prove  that      P  is  the  pole  of  O  ABC. 
Proof.       What  kind  of  angles  are  the  A  A  OP  and  BOP? 
How  is  PO  related  to  the  plane  of  QABC  ? 
Does  this  prove  that  P  is  the  pole  of  QABC  ? 

642.  Describing  Circles  on  a  Sphere.  This  proposition  proves 
that  we  may  describe  a  great  circle  on  a  sphere  of  a  given  radius 
so  that  it  shall  pass  through  two  given  points. 

Open  the  compasses  the  length  of  chord  PA  =  Vr^  +  r'^  =  r  V2, 

643.  Tangent  Lines  and  Planes.  A  line  or  plane  that  has  one 
point  and  only  one  point  in  common  with  a  sphere,  however 
far  produced,  is  said  to  be  tangent  to  the  sphere,  and  the  sphere 
to  be  tangent  to  the  line  or  plane. 

644.  Tangent  Spheres.  Two  spheres  whose  surfaces  have  one 
point  and  only  one  point  in  common  are  said  to  be  tangent. 


386  BOOK  VIII.    SOLID  GEOMETRY 

Proposition  IV.    Theorem: 

645.  A  plane  perpendicular  to  a  radius  at  its  extrem- 
ity is  tangent  to  the  sphere. 


Given  the  plane  MN  perpendicular  to  the  radius  OA  at  A. 

To  prove  that  MN  is  tangent  to  the  sphere. 
Proof.  Let  P  be  any  point  except  A  in  MN. 

Then  which  is  longer,  OP  or  OA,  and  why  ? 
Therefore,  is  P  inside,  on,  or  outside  the  sphere,  and  why  ? 
What  does  this  tell  us   concerning  all   points,    except  A, 
on  MN? 

How,  then,  do  we  know  that  MN  is  tangent  to  the  sphere  ? 

646.  Corollary.  A  plane  tangent  to  a  sphere  is  perpen- 
dicular to  the  radius  drawn  to  the  point  of  contact. 

What  are  the  proposition  and  corollary  of  plane  geometry  corre- 
sponding to  §§  645  and  646  ?    Do  they  suggest  the  proof  of  this  corollary  ? 

647.  Inscribed  Sphere.  If  a  sphere  is  tangent  to  all  the  faces 
of  a  polyhedron,  it  is  said  to  be  inscribed  in  the  polyhedron, 
and  the  polyhedron  to  be  circumscribed  about  the  sphere. 

648.  Circumscribed  Sphere.  If  all  the  vertices  of  a  polyhedron 
lie  on  a  spherical  surface,  the  sphere  is  said  to  be  circumscribed 
about  the  polyhedron,  and  the  polyhedron  to  be  inscribed 
in  the  sphere. 


PLANE  SECTIONS  AND  TANGENT  PLANES    387 

Proposition  V.    Theorem 

649.  A  sphere  may  he  inscribed  in  any  given  tetra- 
hedron. 


Given  the  tetrahedron  ABCD. 

To  prove  that  a  sphere  may  he  inscribed  in  ABCD. 

Proof.  Bisect  the  dihedral  A  at  the  edges  AB,  BC,  and  CA 
by  the  planes  OAB,  OBC,  and  OCA  respectively. 

Every  point  in  the  plane  OAB  is  equidistant  from  the  faces 
ABC  ^nd  ABD.  §479 

Eor  a  like  reason  every  point  in  the  plane  OBC  is  equidistant 
from  the  faces  ABC  and  DBC ;  and  every  point  in  the  plane 
OCA  is  equidistant  from  the  faces  ABC  and  ADC. 

Therefore  the  point  0,  the  common  intersection  of  these 
three  planes,  is  equidistant  from  the  four  faces  of  the  tetra- 
hedron and  is  the  center  of  the  sphere  inscribed  in  the  tetra- 
hedron, by  §  647.  Q.E.D. 

Discussion.  What  is  the  corresponding  proposition  in  plane  geome- 
try ?   Is  the  hne  of  proof  similar  ? 

It  is  shown  in  plane  geometry  that  the  three  lines  which  bisect  the 
three  angles  of  a  triangle  meet  in  a  point.  What  is  the  corresponding 
proposition  with  reference  to  planes  in  a  tetrahedron  ?  Is  it  substan- 
tially proved  in  this  proposition  ? 

It  is  proved  in  plane  geometry  that  a  circle  may  be  inscribed  in  what 
kind  of  a  polygon  ?  What  corresponding  proposition  may  be  inferred  in 
solid  geometry  ? 


388  BOOK  VIII.    SOLID  GEOMETKY 

Proposition  VT.    Theorem 

650.  A  sphere  may  he  circumscribed  about  any  given 
tetrahedron. 


Given  the  tetrahedron  ABCD. 

To  prove  that  a  sphere  may  he  circumscribed  about  ABCD. 

Proof.  Let  P,  Q  respectively  be  the  centers  of  the  circles 
circumscribed  about  the  faces  ABC,  ABD. 

Let  PFi.  be_Lto  the  face  ABC,  and  QS  ±  to  the  face  ABD. 

Then  PR  is  the  locus  of  a  point  equidistant  from  A,  B,  C, 
and  QS  is  the  locus  of  a  point  equidistant  from  A,  B,  D.      §  442 

Therefore  PR  and  QS  lie  in  the  same  plane,  the  plane  ±  to 
AB2it  its  mid-point.  §  443 

If  QS  were  II  to  PR,  it  would  be  ±  to  the  face  ABC.       §  445 

But  this  is  impossible,  for  QS  is  _L  to  the  face  ABD  which 
intersects  the  face  ABC.  Given 

Since  Pit  and  QS  cannot  be  II,  and  since  they  lie  in  the  same 
plane,  they  must  therefore  meet  at  some  point  O. 
.'.  0  is  equidistant  from  A,  B,  C,  and  D, 
and  is  the  center  of  the  required  sphere,  by  §  648.   Q.  e.  d. 

651.  CoKOLLARY.    Through  four  points    not  in   the   same 

plane  one  spherical  surface  and  only  one  can  be  passed. 

The  center  of  any  sphere  whose  surface  passes  through  the  four , 
points  must  be  in  the  planes  mentioned  in  the  proof,  and  since  there  is 
only  one  point  of  intersection,  there  can  be  only  one  sphere. 


PLANE  SECTIONS  AND  TANGENT  PLANES    389 

Pkoposition  VIL    Theorem 

652.  The  intersection  of  two  spherical  surfaces  is  a 
circle  ivhose  plane  is  perj^endicular  to  the  line  which 
joins  the  centers  of  the  spheres  and  whose  center  is  in 
that  line. 


Given  two  intersecting  spherical  surfaces,  with  centers  O  and  O'. 

To  prove  that  the  spherical  surfaces  intersect  in  a  circle 
whose  plane  is  perpendicular  to  00',  and  whose  center  is  in  00'. 

Proof.  Let  the  two  great  circles  formed  by  any  plane 
through  0  and  0'  intersect  in  A  and  B. 

Then  00'  is  a  ±  bisector  of  AB.  §  195 

If  this  plane  revolves  about  00',  the  circles  generate  the 
spherical  surfaces,  and  A  describes  their  line  of  intersection. 

But  during  the  revolution  AC  remains  constant  in  length 
and  perpendicular  to  00'. 

Therefore  .1  generates  a  circle  with  center  C,  whose  plane  is 
perpendicular  to  00',  by  §  432.  q.e.d. 

653.  Spherical  Angle.  The  opening  between  two  great-circle 
arcs  that  intersect  is  called  a  spherical  angle.  A  spherical  angle 
is  considered  equal  to  the  plane  angle  formed  by  the  tangents 
to  the  arcs  at  their  point  of  intersection. 

•     Draw  a  figure  illustrating  this  definition. 

In  elementary  geometry  we  do  not  consider  angles  formed  by  arcs 
of  small  circles. 


390  BOOK  VIIL    SOLID  GEOMETRY 

EXERCISE  101 

1.  The  four  perpendiculars  erected  at  the  centers  of  the 
circles  circumscribed  about  the  faces  of  a  tetrahedron  meet 
in  the  same  point. 

2.  The  six  planes  perpendicular  to  the  edges  of  a  tetra- 
hedron at  their  mid-points  intersect  in  the  same  point. 

3.  The  six  planes  which  bisect  the  six  dihedral  angles  of  a 
tetrahedron  intersect  in  the  same  point. 

4.  Circles  on  the  same  sphere  having  equal  polar  distances 
are  equal. 

5.  Equal  circles  on  the  same  sphere  have  equal  polar  dis- 
tances. 

6.  Find  the  locus  of  a  point  in  a  plane  at  a  given  distance 
from  a  given  point.   Also  of  a  point  in  a  three-dimensional  space. 

7.  A  line  tangent  to  a  great  circle  of  a  sphere  lies  in  the 
plane  tangent  to  the  sphere  at  the  point  of  contact. 

8.  Any  line  in  a  tangent  plane  drawn  through  the  point  of 
contact  is  tangent  to  the  sphere  at  that  point. 

9.  One  plane  and  only  one  plane  can  be  passed  through  a 
given  point  on  a  given  sphere  tangent  to  the  sphere. 

10.  Find  a  point  in  a  plane  equidistant  from  two  intersecting 
lines  in  the  plane,  and  at  a  given  distance  from  a  given  point 
not  in  the  plane.    Discuss  the  solution. 

11.  How  many  points  determine  a  straight  line  ?  a  circle  ? 
a  spherical  surface  ?  Prove  that  two  spherical  surfaces  coin- 
cide if  they  have  this  number  of  points  in  common. 

12.  If  two  planes  which  intersect  in  the  line  AB  touch  a 
sphere  at  the  points  C  and  D  respectively,  the  line  CD  is 
perpendicular  to  AB  in  the  sense  mentioned  in  the  discussion 
under  §  450,  —  that  a  plane  can  be  passed  through  CD  per- 
pendicular to  AB. 


PLANE  SECTIONS  AND  TANGENT  PLANES    391 


Proposition  VIIL    Theorem 

654.  A  sjyherical  angle  is  measured  hy  the  arc  of  the 
great  circle  described  from  its  vertex  as  a  pole  and 
included  between  its  sides,  2)roduced  if  necessary. 


Proof. 


Given  PA  and  PB^  arcs  of  great  circles  intersecting  at  P ;  PA' 
and  PB\  the  tangents  to  these  arcs  at  P;  AB,  the  arc  of  the  great 
circle  described  from  P  as  a  pole  and  included  between  PA  and  PB. 

To  prove  that  the  spherical  Z  APB  is  measured  by  are  AB. 

§185 

§213 

§95 

§461 
§213 


Similarly 
But 


In  the  plane  POB,  PB'  is  _L  to  PO, 
and  OB  is  _L  to  PO. 
.-.  PjB'  is  II  to  OB. 
ZM'is  II  to  OA. 
.'.  Z.A'PB'  =  ZAOB. 
Z  A  OB  is  measured  by  arc  AB. 
.'.  Z  A'PB'  is  measured  by  arc  AB. 
.*.  Z  APB  is  measured  by  arc  yl^,  by  §  653.       q.e.d. 

655.  Corollary  1.  A  spherieal  angle  has  the  same  meas- 
ure as  the  dihedral  angle  formed  hy  the  planes  of  the  two  circles. 

656.  Corollary  2.  All  arcs  of  great  circles  drawn  through 
the  pole  of  a  given  great  circle  are  perpendicular  to  the  given 
circle. 


392  BOOK  VIII.    SOLID  GEOMETRY 

657.  Spherical  Polygon.  A  portion  of  a  spherical  surface 
bounded  by  three  or  more  arcs  of  great  circles  is  called  a 
sjjJierical  'polygon. 

The  bounding  arcs  are  called  the  sides  of  the  polygon,  the  angles 
between  the  sides  are  called  the  angles  of  the  polygon,  and  the  points 
of  intersection  of  the  sides  are  called  the  vertices  of  the  polygon. 

658.  Relation  of  Polygons  to  Polyhedral  Angles.  The  planes 
of  the  sides  of  a  spherical  polygon  form  a  polyhedral  angle 
whose  vertex  is  the  center  of  the  sphere,  whose  face  angles  are 
measured  by  the  sides  of  the  polygon,  and  whose  dihedral  angles 
have  the  same  numerical  measure  as  the  angles  of  the  polygon. 

Thus  the  planes  of  the  sides  of  the  polygon 
ABCD  form  the  polyhedral  angle  0-ABCD. 
The  face  angles  BOA.,  COB.,  and  so  on,  are 
measured  by  the  sides  AB.,  BC,  and  so  on, 
of  the  polygon.  The  dihedral  angle  whose 
edge  is  OA  has  the  same  measure  as  the 
spherical  angle  BAD,  and  so  on. 

Hence  from  any  property  of  polyhedral  angles  ive  may  infer 
an  analogous  property  of  spherical  polygons  ;  and  conversely. 

659.  Convex  Spherical  Polygon.  If  a  polyhedral  angle  at  the 
center  of  a  sphere  is  convex  (§  491),  the  corresponding  spherical 
polygon  is  said  to  be  convex. 

Every  spherical  polygon  is  assumed  to  be  convex  unless  the  contrary 
is  stated. 

660.  Diagonal.  An  arc  of  a  great  circle  joining  two  non- 
consecutive  vertices  of  a  spherical  polygon  is  called  a  diagonal. 

661.  Spherical  Triangle.  A  spherical  polygon  of  three  sides 
is  called  a  spherical  triangle. 

A  spherical  triangle  may  be  right.,  obtuse,  or  acute.  It  may  also  be 
equilateral,  isosceles,  or  scalene. 

662.  Congruent  Spherical  Polygons.  If  two  spherical  polygons 
can  be  applied,  one  to  the  other,  so  as  to  coincide,  they  are  said 
to  be  congiment. 


SPHEKICAL  POLYGONS 
Proposition  IX.    Theorem 


393 


663.  Each  side  of  a  spherical  triangle  is  less  than  the 
sum  of  the  other  tivo  sides. 


Given  a  spherical  triangle  ABC^  CA  being  the  longest  side. 

To  prove  that  CA  <AB  +  BC. 

Proof.    In  the  corresj)onding  trihedral  angle  0-ABC, 

Z  COA  is  less  than  Z  BOA  -\-ZCOB.  §  494 

.-.  CA<AB-i-BC,  hy  §658.  Q.e.d. 


Proposition  X.    Theorem 

664.  The  sum  of  the  sides  of  a  sj^hericcd  polygon  is 
less  than  S60°. 


Given  a  spherical  polygon  ABCD. 

To  prove  that  AB  -hBC-\-  CD  +  DA  <  360°. 

Proof.    In  the  corresponding  polyhedral  angle   0-ABCD, 

Z  BOA  +  Z  COB  +  Z  DOC  +  Z  DOA  <  360°.       §  495 
.'.AB-^  BC  +  CD  -h  DA  <  360°,  by  §  658.  Q.e. d. 


394  BOOK  VIII.    SOLID  GEOMETEY 

665.  Polar  Triangle.  If  from  the  vertices  of  a  spherical  tri- 
angle as  poles  arcs  of  great  circles  are  described,  another 
spherical  triangle  is  formed  which  is  called  the  polar  triangle 
of  the  first. 

Thus,  if  A  is  the  pole  of  the  arc  of  the  great 
circle  B'C\  B  of  C'A%  C  of  A'B',  A'B'C  is  the 
polar  triangle  of  ABC. 

If,  with  ^,  -B,  C  as  poles,  entire  great  circles 
are  described,  these  circles  divide  the  surface  of 
the  sphere  into  eight  spherical  triangles. 

Of  these  eight  triangles,  that  one  is  the  polar 
of  ABC  whose  vertex  A'^  corresponding  to  A^ 
lies  on  the  same  side  of  BC  as  the  vertex  A  ;  and  similarly  for  the  other 
vertices. 

EXERCISE  102 

1.  To  bisect  a  given  great-circle  arc. 
What  must  be  done  to  the  angle  at  the  center  ? 

2.  If  two  great-circle  arcs  intersect,  the  vertical  angles  are 
equal. 

3.  To  describe  an  arc  of  a  great  circle  through  a  given  point 
and  perpendicular  to  a  given  arc  of  a  great  circle. 

4.  Every  point  lying  on  a  great  circle  which  bisects  a  given 
arc  of  another  great  circle  at  right  angles  is  equidistant  (§  635) 
from  the  extremities  of  the  given  arc. 

5.  Two  sides  of  a  spherical  triangle  are  respectively  82° 
47'  and  67°  39'.  What  is  known  concerning  the  number  of 
degrees  in  the  third  side  ? 

6.  Three  sides  of  a  spherical  quadrilateral  are  respectively 
86°  29',  73°  47',  and  69°  54'.  What  is  known  concerning  the 
number  of  degrees  in  the  fourth  side  ? 

7.  Draw  a  picture  of  a  sphere,  and  of  an  equilateral  spherical 
triangle  on  the  sphere,  each  side  being  90°.  Then  draw  a  pic- 
ture of  the  polar  triangle. 


SPHERICAL  POLYGONS 
Proposition  XL    Theorem 


395 


666.  If  one  sj^herical  triangle  is  the  polar  triangle  of 
another,  then  reciprocally  the  second  is  the  polar  tri- 
angle of  the  first. 


Given  the  triangle  ABC  and  its  polar  triangle  A'B'C. 

To  prove  that  ABC  is  the  polar  triangle  of  A'B'C 

Proof.    Since  A  is  the  pole  of  B'C, 

and  C  is  the  pole  of  A'B',  §  665 

.*.  ^'  is  at  a  quadrant's  distance  from  A  and  C.       §  639 
.•.  B'  is  the  pole  of  arc  AC.  .        §  641 

Similarly  A'  is  the  pole  of  BC, 

and  C  is  the  pole  of  AB. 

.-.  ABC  is  the  polar  triangle  of  A'B'C,  by  §  665.        q.e.d. 

Discussion.  Is  it  necessary  that  one  of  the  triangles  should  be  wholly 
within  the  other  ?  Draw  the  figures  approximately,  without  using  instru- 
ments, starting  with  A  ABC  having  AB  =  100°,  AC  =  100°,  BC  =  30°. 

Also  draw  the  figures  having  AB  =  120°,  AC  =  80°,  BC  =  40°. 

Also  draw  the  figures  suggested-  in  Ex.  7,  on  page  394,  where  AB  = 
BC  =  CA  =  90°.     Consider  the  proposition  with  these  figures. 

The  proposition  may  also  be  considered  by  starting  with  A  ABC  as 
the  polar  triangle  of  A  A'B^C,  and  proving  that  A  A'B'C  is  the  polar 
triangle  of  A  ABC. 

It  is  desirable  in  the  study  of  spherical  triangles  to  have  a  spherical 
blackboard.  Where  this  is  not  available,  any  wooden  ball  will  serve  the 
purpose. 


396  BOOK  VIII.    SOLID  GEOMETRY 

Pkoposition  XII.    Theorem 

667.  In  two  jpolar  triangles  each  angle  of  the  one  is 
the  supplement  of  the  opposite  side  in  the  other. 


B'E  =  90°. 
DC  =  90°. 


Given  two  polar  triangles  ABC  and  A'B'C,  the  letter  at  the 
vertex  of  each  angle  denoting  its  value  in  degrees,  and  the  small 
letter  denoting  the  value  of  the  opposite  side  in  degrees. 

To  prove  that  A  +  a' =  1S0°,  B -\-h' =  180°,  C -\- c' =  1S0° ; 

A' -{-a  =180°,^'+^  =180°,  C" 4- 6' =180°. 

Proof.  Produce  the  arcs  AB,  AC  until  they  meet  B^C'  at 
the  pointy  D,  E  respectively. 

Since  B'  is  the  pole  of  AE, 

And  since         C  is  the  pole  of  AD, 

.'.  B'E-}- DC  =1S0°. 
B'D  +  DE-\- DC  =  1S0°, 
DE-{-B'C  =  lSO°. 
But  DE  is  the  measure  of  the  Z.  J., 
B'C  =  a'. 
r.A-\-a'  =  lSO°. 
B-{-b'  =  180°, 
C-\-c'  =  180°. 

In  a  similar  way,  starting  with  A  A' B'C'  and  producing  the 
sides  of  A  ABC,  all  the  other  relations  are  proved.  q.e.d. 


That  is, 
or 

and 

Similarly 
and 


§639 

Ax.  1 
Ax.  9 
Ax.  9 
§654 


SPHEEICAL  POLYGONS 
Proposition  XIII.    Theorem 


397 


668.  The  sum  of  the  angles  of  a  spherical  triangle  is 
greater  than  180°  and  less  than  540°. 


Given  a  spherical  triangle  ABC,  the  letter  at  the  vertex  of  each 
angle  denoting  its  value  in  degrees,  and  the  small  letter  denoting 
the  value  of  the  opposite  side  in  degrees. 

To  prove  that  A -\- B  +  C>  1^0°  and  <  540°. 

Proof.       Let  A  /I '^'6"  be  the  polar  triangle  of  A  ABC. 


Then    A  +  a'  =  180°,  B  +  h'  =  180°,  C  +  c'  =  180^ 
,',  A+B^C-^a'  +  h'  +  c'  =  540°. 
.' ,  A  +  B  +  C  =  M()°  -  (a'  -^h^  +  rJ). 


Now  a^  +  ^'  +  (i^  <  360' 


§667 
Ax.  1 
Ax.  2 
§664 


.'.A+B  +  C 


Q.E.D. 


540°  —  some  value  less  than  360°. 
.•_l+5  +  C'>180°. 
Again  a'  +  &'  +  c'  is  greater  than  0°. 
.-.  ^l+^B+C<540°. 

669.  Corollary.  A  spherical  triangle  may  have  two^  or 
even  three,  right  angles;  and  a  spherical  triangle  mag  have 
two,'  or  even  three,  obtuse  angles. 

670.  Triangles  classified  as  to  Right  Angles.  A  spherical 
triangle  having  two  right  angles  is  said  to  be  blrect angular; 
one  having  three  right  angles  is  said  to  be  trlrectangular. 

The  same  terms  may  be  appHed  to  the  corresponding  trihedral  angles. 


398  BOOK  VIII.    SOLID  GEOMETRY 

EXERCISE  103 

1.  If  two  sides  of  a  spherical  triangle  are  quadrants,  the 
third  side  measures  the  opposite  angle. 

2.  In  a  birectangular  spherical  triangle  the  sides  opposite 
the  right  angles  are  quadrants^,  and  the  side  opposite  the  third 
angle  measures  that  angle. 

Since  the  A  are  rt.  A,  what  two  planes  are  ±  to  a  third  plane  ?  What 
two  arcs  must  therefore  (§  632)  pass  through  the  pole  of  a  third  arc  ? 
Then  what  two  arcs  are  quadrants  ?  Then  how  is  the  third  angle  (§  654) 
measured  ? 

3.  Each  side  of  a  trirectangular  spherical  triangle  is  a 
quadrant. 

4.  Three  planes  passed  through  the  center  of 
a  sphere,  each  perpendicular  to  the  other  two,    c\ 
divide  the  spherical  surface  into  eight  congruent 
trirectangular  triangles.  ^' 

Find  the  number  of  degrees  in  the  sides  of  a  spherical  tri- 
angle, given  the  angles  of  its  polar  triangle  as  follows  : 

5.  82°,  77°,  69°.  8.  83°  40',  48°  57',  103°  43'. 

6.  84|°,  81f  °,  72^°.       9.  96°  37' 40",  82°  29' 30",  68°  47'. 

7.  78°  30',  89°,  102°.   10.  43°  29' 37",  98°  22' 53",  87°  36' 39". 

Find  the  number  of  degrees  in  the  angles  of  a  spherical  tri- 
angle, given  the  sides  of  its  polar  triangle  as  follows  : 

11.  68°  42'  39",  93°  48'  7",  89°  38'  14". 

12.  78°  47'  29",  106°  36'  42",  a  quadrant. 

13.  A  quadrant,  half  a  quadrant,  three  fourths  of  a  quadrant. 

14.  Erom  the  center  of  a  sphere  are  drawn  three  radii,  each 
perpendicular  to  the  other  two.  Find  the  number  of  degrees 
in  the  sides  and  angles  of  the  spherical  triangle  determined 
by  their  extremities. 


SPHERICAL  POLYGONS 


399 


671.  Symmetric  Spherical  Triangles.  If  through  the  center  O 
of  a  sphere  three  diameters  ^yl',  5/^',  CC  are  drawn,  and  the 
points  A,B,  C  are  joined  by  arcs  of  great  circles,  and  also  the 
points  A',B',  C',  the  two  spherical  tri- 
angles ABC  and  A'B'C'  are  called  sv/m- 
metric  spherical  triangles. 

In  the  same  way  we  may  form  two  sym- 
metric polygons  of  any  number  of  sides. 
Having  thus  formed  the  symmetric  polygons, 
we  may  place  them  in  any  position  we  choose 
upon  the  surface  of  the  sphere. 

672.  Relation  of  Symmetric  Triangles.  Two  symmetric  tri- 
angles are  nmtually  equilateral  and  mutually  equiangular ;  yet 
in  general  they  are  not  congruent,  for  they  cannot  be  made  to 
coincide  by  superposition.  If  in  the  above  figure  the  triangle 
ABC  is  made  to  slide  on  the  surface  of  the  sphere  until  the 
vert^  A  falls  on  A ',  it  is  evident  that  the  two  triangles  cannot 
be  made  to  coincide  for  the  reason  that  the  corresponding  parts 
of  the  triangles  occur  in  reverse  order. 

To  try  to  make  two  symmetric  spherical  polygons  coincide  is  very 
much  like  trying  to  put  the  right-hand  glove  on  the  left  hand.  The  rela- 
tion of  two  symmetric  spherical  triangles  may  be  illustrated  by  cutting 
them  out  of  the  peel  of  an  orange  or  an  apple. 

673.  Symmetric  Isosceles  Triangles.  If,  however,  we  have  two 
symmetric  triangles  ABC  and  ^'i^'C  such  that  AB  =  AC,  and 
A'B'  =  A'C',  that  is,  if  the  two  sym- 
metric triangles  are  isosceles,  then 
because  AB,  AC,  A'B',  A'C'  are  all 
equal  and  the  angles  A  and  A'  are 
equal,  being  originally  formed  by 
vertical  dihedral  angles  (§  671),  the  -B 
two  triangles  can  be  made  to  coincide.    Therefore, 

If  tiuo  syw.metric  sjjherical  triangles  are  isosceles,  they  are 
superposahle  and  therefore  are  congruent. 


400  BOOK  VIII.    SOLID  GEOMETRY 

Proposition  XIV.    Theorem 
674.  Tivo  symmetric  spherical  triangles  are  equivalent. 


Given  two  symmetric  spherical  triangles  ABC^  A^B^C\  having 
their  corresponding  vertices  opposite  each  to  each  with  respect  to 
the  center  of  the  sphere. 

To  prove  that  the  triangles  ABC,  A'B'C  are  equivaleiit. 

Proof.  Let  P  be  the  pole  of  a  small  circle  passing  through 
the  points  A,  B,  C,  and  let  POP'  be  a  diameter. 

Draw  the  great-circle  arcsP.4,  PB,  PC,  P'A',  P'B',  P'C. 
Then  PA  =  PB  =  PC.  §  636 

Now  PA'  =  PA,  P'B'  =  PB,  P'C'  =  PC.  §672 

.\P'A'  =  P'B'^P'C'.  Ax.  8 

.\  the  two  symmetric  A  PC  A  and  P'C  A'  are  isosceles. 

.'.  APCA  is  congruent  to  A  P'C 'A'.  §  673 

Similarly         A  PAB  is  congruent  to  A  P'A'B', 
and  A  PBC  is  congruent  to  A  P'B'C. 

Now  A  ABC  =  A  PC  A  +  A  PAB  +  A  PBC, 

and  A  A'B'C  =  A  P'C  A'  +  A  P'A'B'  -f  A  P'B'C.     Ax.  11 

.'.AABCis  equivalent  to  A  ^'^'C,  by  Ax.  9.      q.e.d. 

Discussion.  If  the  pole  P  should  fall  without  the  A  ABC,  then  P' 
would  fall  without  A  A'B'C,  and  each  triangle  would  be  equivalent  to 
the  sum  of  two  symmetric  isosceles  triangles  diminished  by  the  third ;  f,o 
that  the  result  would  be  the  same  as  before. 


SPHERICAL  POLYGONS 
Proposition  XV.    Theorem 


401 


675.  Tivo  triangles  on  the  same  sphere  or  on  equal 
spheres  are  either  congruent  or  symmetric  ifMvo  sides 
and  the  included  angle  of  the  one  are  respectively  equal 
to  the  corresp)onding  parts  of  the  other. 


Given  two  spherical  triangles  ^BCand^'5'C',  ^it\iAB=A^B\ 
AC=:A'C\  and  angle  A  =  angle  A\  and  similarly  arranged ;  and 
given  the  triangle  A^B^X  symmetric  with  respect  to  the  triangle 
A^B^a. 

To  prove  that  A  ABC  is  congruent  to  A  A'B'C,  and  that 
A  ABC  is  symmetric  with  respect  to  A  A'B'X. 

Proof.  Superpose  A  ABC  on  A  A'B'C,  the  proof  being  sim- 
ilar to  that  of  the  corresponding  case  in  plane  geometry.     §  68 

.'.  A  ABC  is  congruent  to  A  A'B'C.  §  662 

Since  A  A'B'X  is  symmetric  with  respect  to  A  A'B'C, 
and  A  ABC  is  congruent  to  A  A'B'C', 

.'.  AC  =  A'X,  AB  =  A'B',  AA=A  XA'B'. 

But  A  ABC  is  congruent  to  A  A'B'C'  and  may  be  made  to 
coincide  with  it. 

.'.A  ABC  is  symmetric  with  respect  to  A  A'B'X.    q.e.d. 

Discussion.  In  the  case  of  plane  triangles,  if  the  corresponding  parts 
are  arranged  in  reverse  order,  we  can  still  prove  the  triangles  congruent. 
Why  can  we  not  do  so  in  the  case  of  spherical  triangles  ? 


402  BOOK  VIII.    SOLID  GEOMETRY 

Proposition  XVI.    Theorem 

676.  Two  triangles  on  the  same  sjphere  or  on  equal 
spheres  are  either  congruent  or  symmetric  if  two  angles 
and  the  included  side  of  the  one  are  respectively  equal 
to  the  corresponding  parts  of  the  other. 


Given  two  spherical  triangles  ABC  and  A^B'C\  with  angle 
^  =  angle  A\  angle  C  =  angle  C,  and  AC  =  A'C\  and  similarly 
arranged  ;  and  given  the  triangle  A'B'X  symmetric  with  respect  to 
the  triangle  A^B'C. 

To  prove  that  A  ABC  is  congruent  to  A  A'B'C,  and  that 
A  ABC  is  symmetric  with  respect  to  AA'B'X. 

Proof.  Superpose  A  ABC  on  AA'B'C',  the  proof  being  simi- 
lar to  that  of  the  corresponding  case  in  plane  geometry.       §  72 

.-.  A  ABC  is  congruent  to  AA'B'C.  §  662 

Since  AA'B'X  is  symmetric  with  respect  to  AA'B'C',  and 
A  ABC  is  congruent  to  A^'^'C, 

.'.ZA  =  ZXA'B',  ZC  =  ZX,  and  AC  =  A'X. 

But  A  ABC  is  congruent  to  AA'B'C  and  may  be  made  to 
coincide  with  it. 

.'.A  ABC  is  symmetric  with  respect  to  A  A'B'X.    q.e.d. 

Discussion.  Under  what  circumstances  are  the  two  triangles  both  con- 
gruent and  symmetric  ? 

In  plane  geometry  what  is  the  case  that  corresponds  to  the  one  in 
which  the  spherical  triangles  are  both  congruent  and  synnnetric  ? 


SPHERICAL  POLYGONS 


403 


Proposition  XVII.    Theorem 

677.  Two  mutually  equilateral  triangles  on  the  same 
sphere  or  on  equal  spheres  are  mutually  equiangular, 
and  are  either  congruent  or  symr}ietric. 


Given  two  spherical  triangles,  ABC^  A'B'C\  on  equal  spheres, 
such  that  AB  =  A'B',  BC  =  B'C\  CA  =  C'A^. 

To  prove  that  ZA^  ZJ^Z  B  =  ZB',  ZC  =  ZC,  and  that 
A  ABC  aiidA'B'C  are  either  congruent  or  symmetric. 

Proof.    Let  0  and  O'  be  the  centers  of  the  spheres. 

Pass  a  plane  through  each  pair  of  vertices  of  each  triangle 

and  the  center  of  its  sphere. 

Then  in  the  trihedral  angles  at  O  and  0'  the  face  angles  are 

equal  each  to  its  corresponding  face  angle.  §  167 

.'.  the  corresponding  dihedral  A  are  respectively  equal.  §  499 
.  • .  the  A  of  the  spherical  A  are  respectively  equal.  §  655 
.'.  the  A  are  either  congruent  or  symmetric,  by  §  676.     q.e.d. 

Discussion.  In  the  figures  the  parts  are  arranged  in  the  same  order, 
so  that  the  triangles  are  congruent.  They  might  be  arranged  as  in  the 
figures  of  §  676. 

Discuss  the  proposition  when  the  triangles  are  equilateral  and  each 
side  is  a  quadrant. 

Discuss  the  proposition  when  two  sides  of  each  triangle  are  quadrants. 

What  is  the  corresponding  proposition  in  plane  geometry,  and  why 
does  not  the  form  of  proof  there  given  hold  here  ? 


404  BOOK  VIII.    SOLID  GEOMETRY 

Proposition  XVIII.    Theorem 

678.  Two  mutually  equiangular  triangles  on  the  same 
sphere  or  on  equal  spheres  are  m^utually  equilateral, 
and  are  either  coiigruent  or  symmetric. 


Given  two  mutually  equiangular  spherical  triangles  T  and  T'  on 
equal  spheres. 

To  prove  that  T  and  T  are  mutually  equilateral^  and  are 
either  congruent  or  symmetric. 

Proof.  Let  the  AP  be  the  polar  triangle  of  A  T^  and  the  AP' 
be  the  polar  triangle  of  A  7". 

Since  the  A  T  and  7"  are  mutually  equiangular,    Given 

.  • .  the  polar  A  P  and  P'  are  mutually  equilateral.      §  667 

.*.  the  polar  AP  and  P'  are  mutually  equiangular.    §  677 

But  the  A  T  and  T'  are  the  polar  A  of  A  P  and  P'.    §  666 

.  • .  the  A  T  and  2"  are  mutually  equilateral.  §  667 

Therefore  the  A  T  and  T  are  either  congruent  or  symmetric, 

by  §  677.  Q.E.D. 

Discussion.  The  statement  that  mutually  equiangular  spherical  tri- 
angles are  mutually  equilateral,  and  are  either  congruent  or  symmetric, 
is  true  only  when  they  are  on  the  same  sphere  or  on  equal  spheres.  When 
the  spheres  are  unequal,  the  spherical  triangles  are  unequal.  In  this  case, 
however,  their  sides  have  the  same  arc  measure,  and  therefore  have  the 
same  ratio  as  the  circumferences  or  as  the  radii  of  the  spheres  (§  382) . 


SPHERICAL  POLYGONS  405 

Proposition  XIX.    Theorem 

679.  In  an  isosceles  spherical  triangle  the  angles  op- 
posite  the  equal  sides  are  equal. 


Given  the  spherical  triangle  ABC,  with  AB  equal  to  AC. 

To  prove  that  AB  =  Z.C. 

Proof.  Draw  the  arc  AD  oi  2i  great  circle,  from  the  vertex  A 
to  the  mid-point  of  the  base  BC. 

Then      A  A  ED  and  ACD  are  mutually  equilateral. 

.'.A  A  ED  and  ^  CD  are  mutually  equiangular.      §  677 
.\AE  =  AC.  q.e.d. 

EXERCISE  104 

1.  The  radius  of  a  sphere  is  4  in.  From  any  point  on  the  sur- 
face as  a  pole  a  circle  is  described  upon  the  sphere  with  an  open- 
ing of  the  compasses  equal  to  3  in.    Find  the  area  of  this  circle. 

2.  The  edge  of  a  regular  tetrahedron  is  a.  Find  the  radii 
r,  r'  of  the  inscribed  and  circumscribed  spheres. 

3.  Find  the  diameter  of  the  section  of  a  sphere  of  diameter 
10  in.  made  by  a  plane  3  in.  from  the  center. 

4.  The  arc  of  a  great  circle  drawn  from  the  vertex  of  an 
isosceles  spherical  triangle  to  the  mid-point  of  the  base  bisects 
the  vertical  angle,  is  perpendicular  to  the  base,  and  divides  the 
triangle  into  two  symmetric  triangles. 


406  BOOK  VIIL    SOLID  GEOMETRY 

Pkoposition  XX.    Theorem 

680.  If  two  angles  of  a  spherical  triangle  are  equal, 
the  sides  opposite  these  angles  are  equal  and  the  tri- 
angle is  isosceles. 


Given  the  spherical  triangle  ABC^  with  angle  B  equal  to  angle  C. 

To  prove  that  AC  =  AB. 

Proof.    Let  A  A'B'C  be  the  polar  triangle  of  A  ABC. 
Since  Z7J  =  ZC,    .'.  A'C  =  A'B'.  §667 

.'.ZB'  =  ZC'.  §679 

.'.  AC  =  AB,  by  §  667.  Q.e.d. 

EXERCISE  105 

1.  To  bisect  a  given  spherical  angle. 

2.  To  construct  a  spherical  triangle,  given  two  sides  and  the 
included  angle, 

3.  To  construct  a  spherical  triangle,  given  two  angles  and  the 
included  side. 

4.  To  construct  a  spherical  triangle,  given  the  three  sides. 

5.  To  construct  a  spherical  triangle,  given  the  three  angles. 

6.  To  pass  a  plane  tangent  to  a  given  sphere  at  a  given  point 
on  the  surface  of  the  sphere. 

7.  To  pass  a  plane  tangent  to  a  given  sphere  through  a  given 
straight  line  without  the  sphere. 


SPHEKICAL  POLYGONS 


407 


Proposition  XXI.    Theorem 

681.  7/"  tiDO  angles  of  a  spherical  triangle  are  unequal^ 
the  sides  opposite  these  angles  are  unequal,  and  the  side 
opposite  the  greater  angle  is  the  greater;  and  if  two  sides 
are  unequal,  the  angles  opposite  these  sides  are  unequal, 
and  the  angle  opposite  the  greater  side  is  the  greater. 


Given  the  triangle  ABC^  with  angle  C  greater  than  angle  B. 

To  prove  that  AB>  AC. 

Proof.    Draw  the  arc  CD  of  a  great  circle,  making  Z.DCB 
equal  to  Z  B.    Then  DB  =  DC.  §  680 

Now  AD-j-DOAC.  §663 

.'.  AD-^DB>AC,  or  AB>AC,  by  Ax.  9.  Q.e.d. 

Given  the  triangle  ABC^  with  AB  greater  than  AC. 

To  prove  that      Z  C  is  greater  than  Z  B. 

Proof.    The  Z  C  must  be  equal  to,  less  than,  or  greater  than 
the  Z  B. 

If  ZC  =  ZB,  then  AB  =  AC',  §  680 

and  if  Z  C  is  less  than  Z B,  then  AB<,AC,SiS  above. 
But  both  of  these  conclusions  are  contrary  to  what  is  given. 
.*.  ZC  is  greater  than  Z.B.  q.e.d. 


408  BOOK  VIII.    SOLID  GEOMETRY 

Propositiok  XXII.    Theorem 

682.  The  shortest  line  that  can  he  draivn  on  the  sur- 
face of  a  sphere  between  tivo  points  is  the  arc  of  a  great 
circle  joining  the  two  points,  not  greater  than  a  semi- 
circle. 


Given  AB^  the  arc  of  a  great  circle,  not  greater  than  a  semicircle, 
joining  the  points  A  and  B. 

To  prove  that  AB  is  the  shortest  line  that  can  he  drawn  on 
the  surface  joining  A  and  B. 

Proof.  Let  C  be  any  point  in  AB. 

With  A  and  B  as  poles  and  A  C  and  BC  as  polar  distances, 
describe  two  arcs  DCF  and  GCE. 

The  arcs  DCF  and  GCE  have  only  the  point  C  in  common. 
For  if  F  is  any  other  point  in  DCF,  and  if  arcs  of  great  circles 
AF  and  BF  are  drawn,  then 

AF=AC.  §636 

But  AF+BF>AC  +  BC.  §663 

Take  away  AF  from  the  left  member  of  the  inequality,  and 
its  equal  A  C  from  the  right  member. 

Then  BF>BC.  Ax.  6 

Therefore  BF>BG,  the  equal  oiBC.        Ax.  9 

Hence  F  lies  outside  the  circle  whose  pole  is  B,  and  the 
arcs  DCF  and  GCE  have  only  the  point  C  in  common. 


SPHERICAL  POLYGONS  409 

Now  let  A  DEB  be  any  line  from  ^  to  ^  on  the  surface  of 
the  sphere,  which  does  not  pass  through  C. 

This  line  will  cut  the  arcs  DCF  and  GCE  in  separate  points 
D  and  E ;  and  if  we  revolve  the  line  A  D  about  A  as  a  fixed 
point  until  D  coincides  with  C,  we  shall  have  a  line  from  A  to 
C  equal  to  the  line  AD. 

In  like  manner,  we  can  draw  a  line  from  J5  to  C  equal  to 
the  line  BE. 

Therefore  a  line  can  be  drawn  from  A  to  B  through  C  that 
is  equal  to  the  sum  of  the  lines  AD  and  BE,  and  hence  is  less 
than  the  line  A  DEB  by  the  line  DE. 

Therefore  no  line  which  does  not  pass  through  C  can  be  the 
shortest  line  from  ^  to  j5. 

Therefore  the  shortest  line  from  A  to  B  passes  through  C. 

But  C  is  any  point  in  the  arc  AB. 

Therefore  the  shortest  line  from  A  to  B  passes  through 
every  point  of  the  arc  AB,  and  consequently  coincides  with 
the  arc  AB. 

Therefore  the  shortest  line  from  A  to  B  is  the  great-circle 
arc  ^5.  Q.E.D. 

EXERCISE  106 

1.  The  three  medians  of  a  spherical  triangle  are  concurrent. 

2.  To  construct  with  a  given  radius  a  spherical  surface  that 
passes  through  three  given  points. 

3.  To  construct  with  a  given  radius  a  spherical  surface  that 
passes  through  two  given  points  and  is  tangent  to  a  given  plane. 

4.  To  construct  with  a  given  radius  a  spherical  surface  that 
passes  through  two  given  points  and  is  tangent  to  a  given 
sphere. 

5.  The  smallest  circle  on  a  given  sphere  whose  plane  passes 
through  a  given  point  within  the  sphere  is  the  circle  whose 
plane  is  perpendicular  to  the  radius  through  the  given  point. 


410 


BOOK  VIII.    SOLID  GEOMETEY 


683.  Zone.  A  portion  of  a  spherical  surface  included  be- 
tween two  parallel  planes  is  called  a  zone. 

Thus  on  the  earth  we  have  the  torrid  zone  included  between  the 
planes  of  the  tropics  of  Cancer  and  Capricorn. 

The  circles  made  by  the  planes  are  called  the 
bases  of  the  zone,  and  the  distance  between  the 
planes  is  called  the  altitiCde  of  the  zone. 

If  one  of  the  planes  is  tangent  to  the  sphere  and 
the  other  plane  cuts  the  sphere,  the  zone  is  called  a 
zone  of  one  base. 

If  both  planes  are  tangent  to  the  sphere,  the  zone 
is  a  complete  spherical  surface. 

684.  Generation  of  a  Zone.  If  a  great  circle  revolves  about 
its  diameter  as  an  axis,  any  arc  of  the  circle  generates  a  zone. 

•  Thus,  in  the  figure  of  §  683,  if  the  great  circle  PACQ  revolves  about 
its  diameter  PQ  as  an  axis,  the  arc  ^C  generates  the  zone  AD,  of  which 
the  altitude  is  the  distance  between  the  parallel  planes.  Similarly,  the 
arc  AP  generates  the  zone  ABP,  and  the  arc  CQ  generates  the  zone  GDQ, 
these  both  being  zones  of  one  base. 

685.  Lune.  A  portion  of  a  spherical  surface  bounded  by 
the  halves  of  two  great  circles  is  called  a  lune. 

Thus  PAQB  is  a  lune.  A  lune  is 
evidently  generated  by  the  partial  or 
complete  revolution  of  half  of  a  great 
circle  about  its  diameter  as  an  axis. 

686.  Angle  of  a  Lune.  The  angle 
between  the  semicircles  bounding 
a  lune  is  called  the  angle  of  the 
lune. 

Thus  /.  APB  is  the  angle  of  the 
lune  PAQB. 

A  lune  is  usually  taken  as  having  an  angle  less  than  a  straight  angle. 
This  is  not  necessary,  for  we  may  consider  a  hemispherical  surface  as  a 
lune  with  an  angle  of  180°.  We  may  also  conceive  of  lunes  with  angles 
greater  than  a  straight  angle,  and  we  may  even  think  of  an  entire 
spherical  surface  as  a  lune  whose  angle  is  360°. 


MEASUREMENT  OF  SPHERICAL  SURFACES    411 

Proposition  XXIII.    Theorem 

687.  The  area  of  the  surface  generated  hy  a  straight 
line  revolving  about  an  axis  in  its  plane  is  equal  to  the 
product  of  the  projection  of  the  liyie  on  the  axis  hy 
the  circle  whose  radius  is  a  perpendicular  erected  at  the 
mid-point  of  the  line  and  terminated  hy  the  axis. 


Given  an  axis  XF  about  which  a  line  AB  in  the  same  plane  with 
XY  revolves,  M  being  the  mid-point  of  AB^  CD  being  the  projec- 
tion of  AB  on  XY^  MO  being  perpendicular  to  XF,  MR  being  per- 
pendicular to  AB^  and  a  being  the  area  generated  by  AB. 

To  prove  that  a=CD  x2  ttMR. 

Proof.  1.  li  AB  is  II  io  XY,  CD=AB,  MR  coincides  with 
MO,  and  a  is  the  lateral  area  of  a  right  cylinder.  §  588 

2.  If  AB  is  not  li  to  XY,  and  does  not  cut  XY,  a  is  the 
lateral  area  of  the  frustum  of  a  cone  of  revolution. 

.\a  =  AB  x27rM0.  §616 

BmwAE  li  to  XY. 

The  AAEB  and  MOR  are  similar.  §  290 

.-.  MO  :  AE  =  MR  :  AB.  §  282 

.'.  AB  X  MO  =  AE  X  MR,  §261 

ov  AB  X  MO  =  CD  X  MR.  Ax.  9 

Substituting,  a=  CD  x   2  ttMR. 

3.  If  A  lies  in  the  axis  XY,  then  AE  and  CD  coincide, 
and                         a  =  CD  x2  it  MR,  by  §  609.  q.e.d. 


412  BOOK  VIII.    SOLID  GEOMETRY 

Proposition  XXIV.    Theorem 

688.  The  area  of  the  surface  of  a  sphere  is  equal  to 
the  j^^oduct  of  the  diameter  by  the  circumference  of  a 
great  circle. 


E     D' 


B'  A 


Given  a  sphere  generated  by  the  semicircle  ABODE  revolving 
about  the  diameter  AE  as  an  axis,  s  being  the  area  of  the  surface, 
r  being  the  radius,  and  d  being  the  diameter. 

To  prove  that  s=2  irrd. 

Proof.  Inscribe  in  the  semicircle  half  of  a  regular  polygon 
having  an  even  number  of  sides,  as  ABODE. 

From  the  center  0  draw  Js  to  the  chords  AB,  BC,  CD,  DE. 
These  Js  bisect  the  chords  (§  174)  and  are  equal.  §  178 

Let  I  denote  the  length  of  each  of  these  Js. 
From  B,  C,  and  D  drop  perpendiculars  to  A  E. 
Then  area  of  surface  generated  hj  AB  =  AB^  x  2  ttI,      §  687 

area  of  surface  generated  hj  BC  =  B'O  x  2  irl,  etc. 
.*.  area  of  surface  generated  by  ABCDE  =  AE  x  2  ttI    Ax.  1 

=  2  irld.  Ax.  9 

Denote  the  area  of  the  surface  generated  by  ABCDE  by  s', 
and  let  the  number  of  sides  of  ABCDE  be  indefinitely  increased. 
Then  s'  approaches  «  as  a  limit, 

I  approaches  r  as  a  limit,  §  377 

and  consequently  2  Trld  approaches  2  irrd  as  a  limit. 

But  s'  =  2  irld,  always.  §  687 

.•.s  =  27rr^,  by  §207.  Q.e.d. 


MEASUEEMENT  OF  SPHERICAL  SURFACES    413 

689.  Corollary  1.  The  area  of  the  surface  of  a  sphere  is 
equivalent  to  the  area  of  four  great  circles^  or  to  4  7rr^ 

In  s  =  2  Trrd,  wliat  is  the  value  of  d  in  terms  of  r  ?  Then  what  is  the 
value  of  s  in  terms  of  r  ? 

For  example,  if  the  radius  is  10  in.,  the  area  of  the  surface  of  the 
sphere  is  47r-  100  sq.  in.,  or  1256.64  sq.  in. 

690.  Corollary  2.    The  areas  of  the  surfaces  of  two  spheres 

are  to  each  other  as  the  squares  on  their  radii,  or  as  the  squares 

on  their  diameters. 

If  the  radii  are  r  and  r%  the  diameters  d  and  d\  and  the  surfaces 
s  and  s',  then  what  is  the  ratio  of  s  to  s',  according  to  §  689  ?  Show  that 
this  also  equals  r^  :  r'^,  and  d^  :  d'^. 

691.  Corollary  3.  The  area  of  a  zone  is  equal  to  the 
product  of  the  altitude  hy  the  circumference  of  a  great  circle. 

If  we  apply  the  reasoning  of  §  688  to  the  zone  generated  by  the  revo- 
lution of  the  arc  JBCJ),  we  obtain 

the  area  of  zone  BCD  =  B'lY  x  2  7rr, 

where  B'B'  is  the  altitude  of  the  zone  and  2  7rr  the  circumference  of 
a  great  circle. 

For  example,  if  the  radius  is  10  in.,  and  the  altitude  is  5  in.,  the  area 
of  the  zone  is  5  •  2  tt  •  10  sq.  in.,  or  314.16  sq.  in. 

692.  Corollary  4.  The  area  of  a  zone  of  one  base  is  equiv- 
alent to  the  area  of  a  circle  whose  radius  is  the  chord  of  the 
generating  arc. 

The  arc  AB  generates  a  zone  of  one  base. 
.-.  the  area  of  the  zone  AB  =  AB'  x  2  7rr  =  ttAB'  x  AE. 
But  •  AW  X  AE  =  AB^.  §  298 

.-.  the  area  of  the  zone  AB  =  ttAB  . 

693.  Spherical  Excess  of  a  Triangle.  The  excess  of  the  sum  of 
the  angles  of  a  spherical  triangle  over  180°  is  called  the  spherical 
excess  of  the  triangle. 

For  example,  if  the  angles  of  a  spherical  triangle  are  80°,  90°,  and  100°, 
the  spherical  excess  of  the  triangle  is  90°. 


414  BOOK  yill.    SOLID  GEOMETRY 

Proposition  XXV.    Theorem 

694.  The  area  of  a  lune  is  to  the  area  of  the  surface  of 
the  sphere  as  the  angle  of  the  lune  is  to  four  right  angles. 


Given  a  lune  PAQB,  the  great  circle  ABCD  whose  pole  is  P,  a 
the  value  in  degrees  of  the  angle  of  the  lune,  /  the  area  of  the  lune, 
and  s  the  area  of  the  surface  of  the  sphere. 

To  prove  that  I :  s  =  a  :  4:  rt.  A. 

Proof.    The  arc  AB  measures  the  Z.a  oi  the  lune.  §  654 

Hence  arc  AB  :  circle  ABCD  =  a  :  4:  rt.  A. 

li  AB  and  ABCD  are  commensurable,  let  their  common  meas- 
ure be  contained  m  times  in  AB,  and  n  times  in  ABCD. 
Then  arc  AB  :  circle  ABCD  =  m  :  n. 

.■.a:4.Tt.A=m:n.  §212 

Pass  an  arc  of  a  great  circle  through  the  poles  P  and  Q  and 
each  point  of  division  of  ABCD. 

These  arcs  will  divide  the  entire  surface  into  n  equal  lunes, 
of  which  the  lune  PA  QB  will  contain  771. 
.' .  I :  s  =  7U  :  n. 
.■ .  I :  s  =  a  :  4  Tt.  A.  Ax.  8 

li  AB  Siiid  ABCD  are  incommensurable,  the  theorem  can  be 
proved  by  the  method  of  limits  as  in  §  472.  q.b.d. 


MEASUREMENT  OF  SPHERICAL  SURFACES    415 

EXERCISE  107 

Using  ir  =  3.1416  for  all  examples  in  this  exercise^  find  the 
areas  of  spheres  whose  radii  are  as  follows : 

1.  2  in.  3.  3^  in.  5.  2  ft.  1  in.  7.  48.8  in. 

2.  7  in.  4.  5|  in.  6.  3  ft.  6  in.  8.  4000  mi. 

Find  the  radii  of  spheres  whose  areas  are  as  follows : 

9.  12.5664  sq.  in.  11.  1  sq.  ft.  13.  s. 

10.  50.2656  sq.  in.  12.  100  tt  sq.  in.  14.  4  ir\ 

On  a  sphere  whose  radius  is  20  in.,  find  the  areas  of  zones 
whose  altitudes  are  as  follows  : 

15.  2  in.  17.  7  in.  19.  1  ft.  21.  3.45  in. 

16.  3  in.  18.  10  in.  20.  2^  in.  22.   6.83  in. 

On  a  sphere  whose  radius  is  10  in.,  find  the  areas  of  luyies 
whose  angles  are  as  follows : 

23.  30°.  25.  90°.  27.  22°  30'.  29.  52°  20'  20". 

24.  45°.  26.  180°.  28.  7°  30'.  30.  48°  35'  10". 

31.  Two  lunes  on  the  same  sphere  or  on  equal  spheres  have 
the  same  ratio  as  their  angles. 

32.  The  area  of  a  lune  is  equal  to  one  ninetieth  of  the  area 
of  a  great  circle  multiplied  by  the  number  of  degrees  in  the 
angle  of  the  lune. 

33.  Zones  on  the  same  sphere  or  on  equal  spheres  are  to 
each  other  as  their  altitudes. 

34.  Given  the  radius  of  a  sphere  15  in.,  find  the  area  of  a 
lune  whose  angle  is  30°. 

35.  Given  the  diameter  of  a  sphere  16  in.,  find  the  area  of  a 
lune  whose  angle  is  75°. 

36.  What  is  the  spherical  excess  of  a  trirectangular  triangle  ? 


416 


BOOK  VIII.    SOLID  GEOMETRY 


Proposition  XXVI.    Theorem 

695.  A  spherical  triangle  is  equivalent  to  a  lune  whose 
angle  is  half  the  sjjherical  excess  of  the  triangle. 


Given  the  spherical  triangle  ABC  on  a  sphere  of  surface  s. 

To  prove  that  A  ABC  is  equivalent  to  a  lune  whose  angle  is 
l(ZA-\-ZB-{-ZC-180°).    . 

Proof.    Produce  the  sides  of  the  A  ABC  to  complete  circles. 

Xow  AAB'C'  and  A'BC  are  symmetric.  Const. 

.'.  A  AB'C  is  equivalent  to  A  A'BC.  §  674 

.'. lune  ABA'C  =  A  ABC -\- AAB'C.  Ax.  9 

But      ACB'A-\-AAC'B-[-AAB'C'-\-AABC  =  ^s.     Ax.  11 

.-.  (lune BCB'A-  A  ABC)  -i-  (lune  CAC'B-AABC) 

-i- lune  ABA'C  =  ^s.  Ax.  9 

.-.  2  A  ABC  =  lune  BCB'A  +  lune  CA  C'B 
+  lune  ABA'C  —  ^s 

.-.A  ABC  =  1  (lune  BCB'A  +  lune  CAC'B 

+  lune  ABA'C 

But  1.  s  =  a  lune  whose  angle  is  180°. 

.*.  A  ^^C  =:  a  lune  whose  angle  is 

^(Z.4+Zi5  +  ZC-180°) 

Discussion.  Since  we  have  found  (§  694)  how  to  compute  the  area  of 
a  lune,  we  can  now  compute  the  area  of  a  spherical  triangle  when  the 
angles  are  known. 


i4 


Axs.  1,  2 


Ax.  4 

§694 


Q.E.D. 


MEASUREMENT  OF  SPHERICAL  SURFACES    417 

696.  Corollary.  If  two  great-circle  arcs  intersect  within 
a  great  circle^  the  sum  of  the  two  opposite  spherical  triangles 
tvhich  they  form  with  the  great  circle  is  equivalent  to  a  lune 
whose  angle  is  the  angle  between  the  arcs. 

697.  Computation  of  Area.  To  illustrate  the  computation  in- 
volved in  §  695,  find  the  area  of  a  triangle  whose  angles  are  110°, 
100°,  and  95°,  on  the  surface  of  a  sphere  whose  radius  is  6  in. 

Spherical  excess  =  110°  +  100°  +  95°  -  180°  =  125°. 

.-.  angle  of  lune  =  62i°. 

62i 
.•.  area  of  lune  =  — -  of  the  spherical  surface. 
360 

621 
.-.  area  of  lune  =  — -  x  4  x  3.1416  x  36  sq.  in. 
,     360 

.'.  area  of  triangle  =  78.54  sq.  in. 

698.  Spherical  Excess  of  a  Polygon.  The  excess  of  the  sum  of 
the  angles  of  a  spherical  polygon  of  n  sides  over  (n  —  2)  x  180° 
is  called  the  spjherical  excess  of  the  polygon. 

EXERCISE  108 

Compute  the  areas  of  triangles  on  spheres  of  the  given  diam- 
eters^ the  angles  being  as  follows  : 

1.  100°,  120°,  140°,  ^  =  16  in.       4.  115°,  124°,  85°,  ^^  =  30  in. 

2.  105°,  130°,  125°,  cZ  =10  in.        5.  135°,  110°,  92°,  6^  =  40  in. 

3.  127°,  132°,  90°,  cZ  =  20  in.        6.  148°,  93°,  68°,  d  =  25.8  in. 

7.  115°  27'  30",  102°  32'  48",  68°  27'  39",  t/=  8000  mi. 

Compute  the  areas  of  triangles  on  spheres  of  the  give?i  radii, 
the  angles  being  as  follows  : 

8.  120°,  100°,  90°,  r  =  9  in.        11.  115°,  102°,  30°,  r  =  36  in. 

9.  130°,  90°,  80°,  r  =  10  in.        12.  140°,  120°,  85°,  r  =  90  in. 
10.  105°,  75°,  65°,  r  =  18  in.        13.  136°,  117°,  93°,  r  =  1.8  in. 


418  BOOK  VIII.    SOLID  GEOMETRY 

Compute  the  areas  of  triangles  on  spheres  of  the  given  cir- 
cu7nferenees,  the  angles  being  as  folloivs : 

14.  93°,  94°,  120°,  c  =  31.416  in. 

15.  82°,  105°,  98°,  G  ==  62.832  in. 

16.  148°,  27°,  125°,  c  =  15.708  in. 

17.  162°,  39°,  120°,  G  =  78.54  in. 

18.  149°,  41°,  116°,  G  =  39.27  in. 

19.  126°  30'  42",  105°  26'  15",  63°  15'  3",  g  =  314.16  in. 

20.  What  is  the  area  of  a  triangle  on  the  earth's  surface 
the  vertices  of  which  are  the  north  pole  and  two  points  on  the 
equator,  one  at  37°  W.  and  the  other  at  16°  E.,  the  earth  being 
considered  a  sphere  with  a  radius  of  4000  mi.  ? 

21.  If  the  radii  of  two  spheres  are  6  in.  and  4  in.  respec- 
tively, and  the  distance  between  the  centers  is  5  in.,  what  is 
the  area  of  the  circle  of  intersection  of  the  spheres  ? 

22.  Eind  the  radius  of  the  circle  determined  on  a  sphere  of 
5  in.  diameter  by  a  plane  1  in.  from  the  center. 

23.  If  the  radii  of  two  concentric  spheres  are  r  and  r',  and 
if  a  plane  is  passed  tangent  to  the  interior  sphere,  what  is  the 
area  of  the  section  made  in  the  other  sphere  ? 

24.  Two  points  A  and  B  are  8  in.  apart.  Eind  the  locus  in 
space  of  a  point  5  in.  from  A  and  7  in.  from  B. 

25.  Two  points  A  and  B  are  10  in.  apart.  Find  the  locus  in 
space  of  a  point  7  in.  from  A  and  3  in.  from  B. 

26.  The  radii  of  two  parallel  sections  of  the  same  sphere  are 
a  and  h  respectively,  and  the  distance  between  the  sections  is 
d.    Eind  the  radius  of  the  sphere. 

27.  The  diameter  of  a  certain  sphere  is  \2.  The  chords  of 
the  arcs  that  form  the  sides  of  a  triangle  on  the  surface  of  the 
sphere  are  respectively  1,  1,  and  I  V2.  Eind  the  area  of  the 
spherical  triangle. 


MEASUREMENT  OF  SPHERICAL  SURFACES    419 

Proposition  XXVII.    Theorem 

699.  A  spherical  polygon  is  equivalent  to  a  lime  whose 
angle  is  half  the  spherical  excess  of  the  polygon. 


Given  a  spherical  polygon  P  of  n  sides,  the  sum  of  the  angles 
being  s. 

To  prove   that  P  is  equivalent  to  a  lune  whose  angle  is 

Proof.    Draw  all  the  diagonals  from  any  vertex. 

Since  there  is  a  distinct  triangle  for  each  side  except  those 
meeting  at  the  vertex  chosen,  there  are  (n  —  2)  triangles. 

Since  each  triangle  is  equivalent  to  a  lune  whose  angle  is 
half  the  excess  of  the  sum  of  its  angles  over  180°,  §  695 

therefore  the  (n  —  2)  triangles  are  equivalent  to  a  lune  whose 
angle  is  half  the  excess  of  the  sum  of  all  the  angles  of  the 
polygon  over  (n  —  2)  X  180°. 

.-.  p  =  a  lune  whose  angle  \^^(s  —  n  —  2  x  180°).     q.e.d. 

700.  Computation  of  Area.  Find  the  area  of  a  spherical  poly- 
gon whose  angles  are  100°,  110°,  120°,  and  170°,  r  being  6  in. 

Spherical  excess  =  100°  +  110°  +  120°  +  170°  -  2  x  180°  =  140°. 
,-.  angle  of  lune  =  70°. 
.-.  area  of  lune  =  ^^g"^  ^'f  4  irr'^ 

=  /e  of  4  X  3.1416  x  36  sq.  in. 

=  87.9648  sq.  in. 


420  BOOK  VIII.    SOLID  GEOMETRY 

EXERCISE  109 

Find  the  areas  of  spherical  polygons  on  spheres  of  the  given 
areas,  the  angles  being  as  follows : 

1.  30°,  90°,  120°,  130°,  a  =2  sq.  ft. 

2.  45°,  60°,  100°,  165°,  a  =  288  sq.  in. 

3.  70°,  168°,  92°,  120°,  a  =  500  sq.  in. 

4.  68°  30',  149°  50',  96°  54',  136°  52',  a  =  750  sq.  in. 

5.  122°27'40",  130°32'50",  98°31'30",  96°48',  a  =  600sq.in. 

6.  132°,  96°,  154°,  120°,  150°,  a  =  3sq.  ft.  120  sq.  in. 

7.  130°,  156°,  172°,  95°,  120°,  100°,  a  =  157.2  sq.  in. 

Find  the  areas  of  spherical  polygons  on  spheres  of  the  given 
radii,  the  angles  being  as  follows: 

8.  130°,  150°,  80°,  90°,  r  =  10  in. 

9.  148°,  157°,  90°,  100°,  120°,  r  =  20  in. 

10.  172°,  169°,  86°,  141°,  100°,  90°,  r  =  24  in. 

11.  135°  30',  148°  42',  96°  37',  102°  11',  r  =  10  in. 

Find  the  areas  of  spherical  polygons  on  spheres  of  the  given 
diameters,  the  angles  being  as  follows : 

12.  148°,  92°,  60°,  120°,  d  =  10  in. 

13.  172°,  168°,  93°,  37°,  100°,  ^  =  22  in. 

14.  102°,  162°,  139°,  141°,  138°,  126°,  d  =  20  in. 

15.  82°50'42",  120°  29'  18",  98°37'15",  141°22'45",  d=20  in. 

Find  the  areas  of  spherical  polygons  on  spheres  of  the  given 
circumferences,  the  angles  being  as  follows : 

16.  39°,  148°,  172°,  168°,  c  =  3.1416  in. 

17.  128°,  92°,  168°,  109°,  c  =  31.416  in. 

18.  146°,  129°,  102°,  137°,  100°,  c  =  6.2832  in. 

19.  128°,  145^  139°,  82°,  161°,  137°,  c  =  18.8496  in. 


MEASUREMENT  OF  SPHERICAL  SOLIDS      421 


701.  Spherical  Pyramid.    A  portion  of  a  sphere  bounded  by 
a  spherical  polygon  and  the  planes  of  its 
sides  is  called  a  spherical  pyramid. 

The  center  of  the  sphere  is  called  the  vertex 
of  the  spherical  pyramid,  and  the  spherical 
polygon  is  called  the  base. 

Thus  0-ABC  is  a  spherical  pyramid. 

702.  Spherical  Sector.  A  portion  of  a  sphere  generated  by 
the  revolution  of  a  circular  sector  about  any  diameter  of  the 
circle  of  which  the  sector  is  a  part  is  called  a  spherical  sector. 

M 


Thus  if  the  sector  A  OB  revolves  about  the  diameter  MN  as  an  axis,  it 
generates  the  spherical  sector  AB-O-A'B'. 

The  zone  generated  by  the  arc  of  the  generating  sector  is  called  the 
base  of  the  spherical  sector, 

703.  Spherical  Segment.  A  portion  of  a  sphere  contained 
between  two  parallel  planes  is  called  a  spherical  segment. 

The  sections  of  the  sphere  made  by  the  parallel  planes  are  called  the 
bases  of  the  spherical  segment,  and  the  distance  between  these  bases  is 
called  the  altitude  of  the  spherical  segment. 

If  one  of  the  parallel  planes  is  tangent  to  the  sphere,  the  segment 
is  called  a  spherical  segment  of  one  base. 

A  spherical  segment  of  one  base  may  be  generated  by  the  revolution 
of  a  circular  segment  about  the  diameter  perpendicular  to  its  base. 

704.  Spherical  Wedge.  A  portion  of  a  sphere  bounded  by  a 
lune  and  the  planes  of  two  great  circles  is  called  a  spherical 
wedge. 


422  BOOK  VIIL    SOLID  GEOMETRY 

Proposition  XXVIII.    Theorem 

705.  Tlie  volume  of  a  sphere  is  equal  to  the  product 
of  the  area  of  its  surface  by  one  third  of  its  radius. 


Given  a  sphere  of  radius  r,  area  of  surface  s,  volume  y,  and 
center  O. 


X 


To  prove  that  v 

Proof.  We  may  imagine  a  cube  of  edge  2  7'  circumscribed 
about  the  sphere. 

Connect  0  with  each  of  the  vertices  of  this  cube. 

These  connecting  lines  are  the  edges  of  six  pyramids  whose 
bases  are  the  faces  of  the  cube  and  whose  altitudes  all  equal  r. 

The  volume  of  each  pyramid  is  a  face  of  the  cube  multiplied 
by  ^r,  and  the  volume  of  the  six  pyramids,  or  of  the  whole 
cube,  is  the  area  of  the  surface  of  the  cube  multiplied  by  ^  r. 

Now  imagine  planes  drawn  tangent  to  the  sphere,  at  the 
points  where  the  edges  of  the  pyramids  cut  its  surface.  We 
then  have  a  circumscribed  solid  whose  volume  is  nearer  that 
of  the  sphere  than  is  the  volume  of  the  circumscribed  cube, 
but  is  still  greater  than  'the  sphere.  Ax.  11 

Proceeding  as  before,  connect  0  with  the  vertices  of  the 
new  polyhedron.  These  connecting  lines  are  the  edges  of 
pyramids  whose  bases  are  together  equal  to  the  bases  of  the 
polyhedron  and  whose  common  altitude  is  r.  §  646 


MEASUREMENT  OF  SPHERICAL  SOLIDS     428 

Then  the  sum  of  the  volumes  of  these  pyramids  is  again  the 
area  of  the  surface  of  the  polyhedron  multiplied  by  ^  r.  De- 
noting this  volume  by  v^  and  the  area  of  the  surface  by  s\  we 
have  ^f^^^X^r. 

If  we  continue  to  draw  tangent  planes  to  the  sphere,  we  con- 
tinue to  diminish  the  circumscribed  solid. 

By  continuing  this  process  indefinitely  we  can  make  the 
difference  between  the  volume  of  the  sphere  and  the  volume 
of  the  circumscribed  solid  less  than  any  assigned  positive 
quantity,  however  small,  the  difference  between  the  surface  of 
the  sphere  and  the  surface  of  the  circumscribed  solid  becoming 
and  remaining  less  than  any  assigned  value,  however  small. 

.•.  V  is  the  limit  of  v',  and  s  is  the  limit  of  s\       §  204 
And  since  it  has  been  shown  that 

v'  =  s'  X  \r,  always, 
.•.v  =  sx\r,hj^201.  Q.E.D. 

706.  Corollary  1.    The  volume  of  a  sphere  of  radius  r  and 

diameter  d  is  equal  to  ^  irr^  or  1  ird^. 

For  in  V  =  s  X  1^  r  what  is  the  value  of  s  in  terms  of  r  ?  What  is  the 
value  of  d  in  terms  of  r  ?   Then  what  is  the  value  of  v  in  terms  of  d? 

707.  Corollary  2.    The  volumes  of  two  spheres  are  to  each 

other  as  the  cubes  of  their  radii. 

What  is  the  ratio  of  |  irr^  to  |  irr'^  ? 

By  tlie  same  reasoning,  tlie  volumes  are  to  each  other  as  the  cubes  of 
the  diameters. 

708.  Corollary  3.    The   volume   of  a   spherical  sector  is 

equal  to  one  third  the  product  of  the  area  of  the  zone  which 

forms  its  base  multiplied  by  the  radius  of  the  sphere. 

Suppose  the  base  divided  into  spherical  triangles.  The  planes  deter- 
mined by  their  vertices  are  the  bases  of  triangular  pyramids  with  ver- 
tices at  0.  What  is  the  limit  of  the  sum  of  the  volumes  of  these  pyramids 
as  the  bases  decrease  in  size  ? 


424  BOOK  VIII.    SOLID  GEOMETRY 

EXERCISE  110 
Problems  of  Computation 
Find  the  volumes  of  spheres  whose  radii  are  : 

1.  3  in.  4.  21  in.  7.  20.7  ft. 

2.  5  in.  5.  4|  in.  8.  2  ft.  3  in. 

3.  7  in.  6.  9^  in.  9.  4000  mi. 

Find  the  volumes  of  spheres  whose  diameters  are : 

10.  24  in.  13.  2.8  in.  16.  2  ft.  1  in. 

11.  36  in.  14.  3.4  in.  17.  3  ft.  4  in. 

12.  48  in.  15.  4.5  in.  18.  8  ft.  6  in. 

Find  the  volumes  of  spheres  whose  circumferences  are  : 
19.  6.2832  in.  20.  12.5664  in.  21.  18.8496  in. 

Find  the  volumes  of  spheres  whose  surface  areas  are  : 

22.  12.5664  sq.  in.    23.  50.2656  sq.  in.    24.  113.0976  sq.  in. 

Find  the  radii  of  spheres  whose  volumes  are : 

25.  4.1888  cu.  in.      26.  33.5104  cu.  in.    27.  .113.0976  cu.  in. 

28.  The  circumference  of  a  hemispherical  dome  is  66  ft. 
How  many  square  feet  of  lead  are  required  to  cover  it  ? 

29.  If  the  ball  on  the  top  of  St.  Paul's  Cathedral  in  London 
is  6  ft.  in  diameter,  how  much  would  it  cost  to  gild  it  at  9  cents 
per  square  inch  ? 

30.  The  dihedral  angles  made  by  the  faces  of  a  spherical 
pyramid  are  80°,  100°,  120°,  and  150°,  and  the  length  of  a 
lateral  edge  is  42  ft.    Find  the  area  of  the  base. 

31.  The  dihedral  angles  made  by  the  faces  of  a  spherical 
pyramid  are  60°,  80°,  and  100°,  and  the  area  of  the  base  is 
4  TT  sq.  ft.    Find  the  radius. 


EXERCISES  425 

32.  What  is  the  area  of  the  surface  of  the  earth  ? 

Assume  that  the  earth  is  a  sphere  with  a  radius  of  4000  mi.,  and  make 
the  same  assumption  in  subsequent  examples  relating  to  the  earth. 

33.  The  altitude  of  the  torrid  zone  is  3200  mi.   Find  its  area. 

34.  What  is  the  area  of  the  north  temperate  zone  if  its 
altitude  is  1800  mi.  ? 

35.  Find  the  number  of  square  miles  of  the  earth's  surface 
that  can  be  seen  from  an  aeroplane  1500  ft.  above  the  surface. 

36.  How  far  in  one  direction  can  a  man  see  from  the  deck 
of  an  ocean  steamer  if  his  eye  is  40  ft.  above  the  water  ? 

37.  To  what  height  must  a  man  be  raised  above  the  earth 
in  order  to  see  one  sixth  of  its  surface  ? 

38.  How  much  of  the  earth's  surface  would  a  man  see  if  he 
were  raised  to  the  height  of  the  radius  above  it  ? 

39.  If  the  atmosphere  extends  50  mi.  above  the  surface  of 
the  earth,  find  the  volume  of  the  atmosphere. 

40.  If  an  iron  ball  4  in.  in  diameter  weighs  9  lb.,  find  the 
weight  of  a  spherical  iron  shell  2  in.  thick,  the  external  diame- 
ter being  20  in. 

41.  What  is  the  angle  of  a  spherical  wedge  if  its  volume  is 
11  cu.  ft.  and  the  volume  of  the  entire  sphere  is  8|  cu.  ft.  ? 

42.  The  inside  of  a  washbasin  is  in  the  shape  of  the  segment 
of  a  sphere.  The  distance  across  the  top  is  16  in.  and  its 
greatest  depth  is  6  in.  How  many  pints  of  water  will  it  hold, 
allowing  7  gal.  to  the  cubic  foot  ? 

43.  Prove  that  the  volume  of  a  spherical  pyramid  is  equal 
to  the  product  of  the  base  by  one  third  of  the  radius,  and  find 
the  volume  if  the  base  is  one  eighth  of  the  surface  of  a  sphere 
of  radius  10  in. 

44.  Find  the  volume  of  a  spherical  sector  whose  base  is  a 
zone  of  area  z^  the  radius  of  the  sphere  being  /-,  following  a 
process  of  reasoning  similar  to  that  in  §  705. 


426  BOOK  VIII.    SOLID  GEOMETRY 

EXERCISE  111 
Formulas 

1.  Find  the  area  z  of  the  zone  of  a  sphere  of  radius  r,  illu- 
minated by  a  lamp  placed  at  the  height  h  above  the  surface. 

2.  Find  the  volume  v  of  a  sphere  in  terms  of  c,  the  cir- 
cumference. 

3.  Find  the  radius  r  of  a  sphere  in  terms  of  v,  the  volume. 

4.  Find  the  diameter  c?  of  a  sphere  in  terms  of  s^  the 
area  of  the  surface. 

5.  Find  the  circumference  c  of  a  sphere  in  terms  of  s,  the 
area  of  the  surface. 

6.  What  is  the  altitude  a  of  a  zone,  if  its  area  is  z  and  the 
volume  of  the  sphere  is  v  ? 

7.  Show  that  in  a  spherical  pyramid  v  =  \  hr.  Find  r  in 
terms  of  v  and  h ;  also  h  in  terms  of  v  and  r. 

8.  Find  a  formula  for  the  volume  of  the  metal  in  a  spher- 
ical iron  shell,  the  inside  radius  being  r  and  the  thickness  of 
the  metal  being  t. 

9.  Find  a  formula  for  the  weight  of  a  spherical  shell,  the 
inside  radius  being  r,  the  thickness  of  the  metal  being  t,  and 
the  weight  of  a  cubic  unit  of  metal  being  w. 

10.  If  the  area  of  a  zone  z  equals  2  Trra  (§  691),  find  a  for- 
mula for  a  in  terms  of  z  and  r. 

11.  If  the  area  of  a  zone  is  expressed  by  the  formula  z  =  2  irra, 
what  is  the  diameter  of  the  sphere  upon  which  a  zone  z  has  an 
altitude  a  ? 

12.  Find  the  area  ^  of  a  zone  of  altitude  a  on  a  sphere  whose 
area  of  surface  is  s. 

13.  Find  a  formula  for  the  area  a  of  that  part  of  the  surface 
of  a  sphere  of  radius  r  seen  from  a  point  at  a  distance  d  above 
the  surface. 


EXERCISES  427 

EXERCISE  112 
Problems  of  Loci 
Find  the  locus  of  a  point  : 

1.  At  a  given  distance  from  a  given  point. 

2.  At  a  given  distance  from  a  given  straight  line. 

3.  At  a  given  distance  from  a  given  plane. 

4.  At  a  given  distance  from  a  given  cylindric  surface. 

5.  At  a  given  distance  from  a  given  spherical  surface. 

6.  Equidistant  from  two  given  points. 

7.  Equidistant  from  two  given  planes. 

8.  At  a  given  distance  from  a  given  point  and  at  another 
given  distance  from  a  given  straight  line. 

9.  At  a  given  distance  from  a  given  point  and  at  another 
given  distance  from  a  given  plane. 

10.  At  a  given  distance  from  a  given  point  and  equidistant 
from  two  other  given  points. 

11.  At  a  given  distance  from  a  given  point  and  equidistant 
from  two  given  planes. 

Find  one  or  more  points  : 

12.  At  a  distance  d^  from  a  given  point,  at  a  distance  d^  from 
a  given  straight  line,  and  at  a  distance  d^  from  a  given  plane. 

13.  At  a  distance  d^  from  a  given  point,  at  a  distance  d^ 
from  a  given  plane,  and  equidistant  from  two  other  given 
planes. 

14.  Equidistant  from  two  given  points,  equidistant  from  two 
given  planes,  and  at  a  distance  r  from  a  given  point. 

15.  Find  the  locus  of  the  center  of  a  sphere  whose  surface 
touches  two  given  planes  and  passes  through  two  given  points 
that  lie  between  the  planes. 


428  BOOK  VIIL    SOLID  GEOMETRY 

EXERCISE  113 
Miscellaneous  Exercises 

1.  The  volume  of  a  sphere  is  to  the  volume  of  the  inscribed 
cube  as  tt  is  to  |  V3. 

2.  The  volume  of  a  sphere  is  to  the  volume  of  the  circum- 
scribed cube  as  it  is  to  6. 

3.  Find  the  ratio  of  the  volume  of  a  cube  inscribed  in  a 
sphere  to  that  of  a  circumscribed  cube. 

4.  Find  the  difference  between  the  volumes  of  two  cubes, 
one  inscribed  in  a  sphere  of  radius  10  in.  and  the  other  circum- 
scribed about  it. 

5.  The  planes  perpendicular  to  the  three  faces  of  a  trihedral 
angle,  and  bisecting  the  face  angles,  meet  in  a  straight  line. 

6.  The  planes  that  pass  through  the  edges  of  a  trihedral 
angle,  and  are  perpendicular  to  the  opposite  faces,  meet  in  a 
straight  line. 

7.  The  altitude  of  a  regular  tetrahedron  is  equal  to  the  sum 
of  four  perpendiculars  let  fall  from  any  point  within  the  tetra- 
hedron upon  the  four  faces. 

8.  To  cut  a  given  tetrahedral  angle  by  a  plane  so  that  the 
section  shall  be  a  parallelogram. 

9.  Compare  the  volumes  of  the  solids  generated  by  the 
revolution  of  a  rectangle  successively  about  two  adjacent  sides, 
the  sides  being  a  and  h  respectively. 

10.  Find  the  difference  between  the  volume  of  a  frustum  of 
a  pyramid  and  the  volume  of  a  prism  each  24  ft.  high,  if  the 
bases  of  the  frustum  are  squares  with  sides  20  ft.  and  16  ft. 
respectively,  and  the  base  of  the  prism  is  the  section  of  the 
frustum  parallel  to  the  bases  and  midway  between  them. 

11.  To  draw  a  line  through  the  vertex  of  any  trihedral  angle, 
making  equal  angles  with  its  edges. 


EXERCISES  429 

12.  The  lines  drawn  from  each  vertex  of  a  tetrahedron  to 
the  point  of  intersection  of  the  medians  of  the  opposite  face  all 
meet  in  a  point  called  the  center  of  gravity  of  the  tetrahedron, 
which  divides  each  line  so  that  the  ratio  of  the  shorter  seg- 
ment to  the  whole  line  is  1 : 4. 

13.  The  lines  joining  the  mid-points  of  the  opposite  edges 
of  a  tetrahedron  all  pass  through  the  center  of  gravity  and  are 
bisected  by  it. 

14.  The  plane  which  bisects  a  dihedral  angle  of  a  tetrahe- 
dron divides  the  opposite  edge  into  segments  proportionaL  to 
the  areas  of  the  faces  that  include  the  dihedral  angle. 

15.  To  cut  a  given  cube  by  a  plane  so  that  the  section  shall 
be  a  regular  hexagon. 

16.  The  volume  of  a  right  circular  cylinder  is  equal  to  the 
product  of  the  lateral  area  by  half  the  radius. 

17.  The  volume  of  a  right  circular  cylinder  is  equal  to  the 
product  of  the  area  of  the  rectangle  which  generates  it,  by  the 
length  of  the  circumference  generated  by  the  point  of  intersec- 
tion of  the  diagonals  of  the  rectangle. 

18.  If  the  altitude  of  a  right  circular  cylinder  is  equal  to 
the  diameter  of  the  base,  the  volume  is  equal  to  the  total  area 
multiplied  by  a  third  of  the  radius. 

19.  The  surface  of  a  sphere  is  two  thirds  the  total  surface 
of  the  circumscribed  cylinder. 

20.  The  volume  of  a  sphere  is  two  thirds  the  volume  of  the 
circumscribed  cylinder. 

21.  Given  a  sphere,  a  cylinder  circumscribed  about  the 
sphere,  and  a  cone  of  two  nappes  inscribed  in  the  cylinder. 
If  any  two  planes  are  drawn  perpendicular  to  the  axis  of  the 
three  figures,  the  spherical  segment  between  the  planes  is 
equivalent  to  the  difference  between  the  corresponding  cylin- 
dric  and  conic  segments.  ^ 


430  BOOK  VIII.    SOLID  GEOMETEY 

EXERCISE  114 

Eeview  Questions 

1.  How  is  a  sphere  generated  ? 

2.  What  are  two  tests  of  equality  of  spheres  ? 

3.  If  a  plane  cuts  a  sphere,  what  figure  is  formed  ?  Is  the 
same  true  of  a  plane  cutting  a  cone  ? 

4.  What  is  the  test  of  equal  circles  on  a  given  sphere  ? 

5.  What  is  a  great  circle  of  a  sphere  ?  Name  four  proper- 
ties of  great  circles. 

6.  What  is  meant  by  a  plane  being  tangent  to  a  sphere  ? 
State  any  proposition  concerning  a  tangent  plane,  and  the  cor- 
responding proposition  in  plane  geometry. 

7.  Complete  this  statement :  A  sphere  may  be  inscribed 
in  •  •  • .    State  the  corresponding  proposition  in  plane  geometry. 

8.  Complete  this  statement :  A  sphere  may  be  circum- 
scribed about  •  •  •.  State  the  corresponding  proposition  in 
plane  geometry. 

9.  Complete  this  statement :  A  spherical  surface  is  deter- 
mined by  •  •  •  points  not  in  the  same  plane.  State  the  corre- 
sponding proposition  in  plane  geometry. 

10.  What  is  the  limit  of  the  sum  of  the  sides  of  a  spherical 
polygon  ?  What  are  the  limits  of  the  sum  of  the  angles  of  a 
spherical  triangle  ? 

11.  What  is  a  polar  triangle?  State  two  propositions  re- 
lating to  polar  triangles. 

12.  What  is  meant  by  symmetric  spherical  triangles  ?  State 
two  propositions  relating  to  such  triangles. 

13.  State  two  propositions  relating  to  congruent  spherical 
triangles. 

14.  How  is  the  area  of  a  spherical  triangle  found  ?  How  is 
the  area  of  a  spherical  polygon  found  ? 


APPENDIX 


709.  Subjects  Treated.  As  with  plane  geometry,  so  with 
solid  geometry,  there  are  many  topics  that  might  be  taken  in 
addition  to  those  given  in  any  textbook.  The  theorems  and 
problems  already  given  in  this  work  are  standard  propositions 
that  are  looked  upon  as  basal,  and  are  usually  required  as 
preliminary  to  more  advanced  work,  and  these,  with  a  reason- 
able selection  from  the  exercises,  will  be  all  that  most  schools 
have  time  to  consider.  It  occasionally  happens,  however,  that 
a  school  is  able  to  do  more  than  this,  and  then  more  exercises 
may  be  selected  from  the  large  number  contained  in  this  work, 
and  a  few  additional  topics  may  be  studied.  For  this  latter 
purpose  the  appendix  is  added,  but  its  study  should  not  be 
undertaken  at  the  expense  of  good  work  on  the  fundamental 
propositions  and  the  exercises  depending  upon  them. 

The  subjects  treated  are  certain  additional  propositions  in 
the  mensuration  of  solids,  and  a  few  general  theorems  relating 
to  similar  polyhedrons,  these  being  occasionally  required  for 
college  examinations.  There  is  also  added  a  brief  sketch  of  the 
history  of  geometry,  which  all  students  are  advised  to  read  as 
a  matter  of  general  information,  and  a  few  of  those  recreations 
of  geometry  that  add  a  peculiar  interest  to  the  subject. 

710.  Similar  Polyhedrons.  Polyhedrons  that  have  the  same 
number  of  faces,  respectively  similar  and  similarly  placed,  and 
their  corresponding  polyhedral  angles  equal,  are  called  similar 
polyhedrons. 

.    It  will  be  seen  that  this  is  analogous  to  the  definition  of  similar 
polygons  in  plane  geometry. 

431 


432 


APPENDIX  TO  SOLID  GEOMETEY 


Proposition  I.    Theorem 


711.  A  truncated  triangular  prism  is  equivalent  to 
the  sum  of  three  pyramids  whose  common  hase  is  the 
base  of  the  prism  and  whose  vertices  are  the  three  ver- 
tices of  the  inclined  section. 


Given  a  truncated  triangular  prism  ABC-DEF  whose  base  is 
ABC  and  inclined  section  DEF^  the  truncated  prism  being  divided 
into  the  three  pyramids  E-ABC,  E-ACD,  and  E-CFD. 

To  prove  ABC-DEF  equivalent  to  the  sum  of  the  three  pyr- 
amids E-ABC,  D-ABC,  and  F-ABC. 

Proof.    E-ABC  has  the  base  ABC  and  the  vertex  E. 

Now  pyramid  E-A  CD  =  pyramid  B-A  CD.  §  558 

{For  they  have  the  same  hase,  A  CD,  and  the  same  altitude,  since  their 
vertices  E  and  B  are  in  the  line  EB  \\  to  the  plane  ACD.) 

But  the  pyramid  B-A  CD  may  be  regarded  as  having  the  base 
ABC  and  the  vertex  D;  that  is,  as  pyramid  D-ABC. 


PEISMS 


433 


Then  since  A  CFD  and  A  CF  have  the  common  base  CF  and 
equal  altitudes,  their  vertices  lying  in  the  line  AD  which  is 
parallel  to  CF,  they  are  equivalent.  §  326 

Furthermore,  pyramids  E-CFD  and  B-A  CF  not  only  have 
equivalent  bases,  the  A  CFD  and  A  CF,  but  they  have  the  same 
altitude,  since  their  vertices  E  and  B  are  in  the  line  EB  which 
is  parallel  to  the  plane  of  their  bases. 

.*.  pyramid  E-CFD  =  pyramid  B-ACF.  §  558 

But  the  pyramid  B-A  CF  may  be  regarded  as  having  the  base 
ABC  and  the  vertex  F;  that  is,  as  pyramid  F-ABC. 

Therefore  the  truncated  triangular  prism  ABC -DEE  is 
equivalent  to  the  sum  of  the  three  pyramids  E-ABC,  D-ABC, 
and  F-ABC.  q.e.d. 


712.  Corollary  1.  The  volume  of  a  truncated  right  tri- 
angular prism  is  equal  to  the  product  of  its  base  by  one  third 
the  sum  of  its  lateral  edges. 

For  the  lateral  edges  DJ.,  EB,  FC  (Fig.  1),  being  perpendicular  to 
the  base  ABC,  are  the  altitudes  of  the  three  pyramids  whose  sum  is 
equivalent  to  the  truncated  prism.  It  is  interesting  to  consider  the  spe- 
cial case  in  which  A  DEF  is  parallel  to  A  ABC. 

713.  Corollary  2.  The  volume  of  any  truncated  triangular 
prism  is  equal  to  the  product  of  its  right  section  by  one  third 
the  sum  of  its  lateral  edges. 

Eor  the  right  section  DEF  (Fig.  2)  divides  the  truncated  prism  into 
two  truncated  right  prisms. 


434 


APPENDIX  TO  SOLID  GEOMETRY 


Proposition  II.    Theorem 


714.  The  volumes  of  two  tetrahedrons  that  have  a 
trihedral  angle  of  the  one  equal  to  a  trihedral  angle 
of  the  other  are  to  each  other  as  the  products  of  the 
three  edges  of  these  trihedral  angles. 


Given  the  two  tetrahedrons  S-ABC  and  S'-^'5'C',  having  the 
trihedral  angles  S  and  S'  equal,  v  and  u'  denoting  the  volumes. 

To  prove  that       -  =  s,^,  ^  ^'B' x  S'C' 

Proof.    Place  the  tetrahedron  S-ABC  upon  S'-A'B'C  so  that 
the  trihedral  Z  S  shall  coincide  with  the  equal  trihedral  ZS'. 
Draw  CD  and  CD'  _L  to  the  plane  S'A'B', 
and  let  their  plane  intersect  S'A'B'  in  S'DD'. 

The  faces  S'AB  and  S'A'B'  may  be  taken  as  the  bases,  and 
CD,  CD'  as  the  altitudes,  of  the  triangular  pyramids  C-S'AB 
and  C'-S'A'B'  respectively. 

§562 
§332 

§282 


Then 

V 

v'^ 

S'AB  X  CD         S'AB  ^    CD 

S'A'B' 

X  CD'       S'A'B'  '^  CD' 

But 

S'AB         S'A  X  S'B 
S'A'B' ~  S'A'  X  S'B'' 

and 

CD        S'C 
CD'       S'C 

V 

S'A 

X  S'B  X  S'C 

SA  X  SB  X  SC 

"  v'~ 

S'A' 

X  S'B 

xs'c 

~~  S'A'  X  S'B'  X  S'C' 

-J  by  Ax.  9.   Q.E.D. 


POLYHEDRONS  435 

Proposition  III.    Theorem 

715.  In  any  polyhedron  the  number  of  edges  increased 
by  two  is  equal  to  the  number  of  vertices  increased  by 
the  numher  of  faces. 


Given  the  polyhedron  AG,  e  denoting  the  number  of  edges,  v  the 
number  of  vertices,  and  /  the  number  of  faces. 

To  prove  that  e  -{-2  =  v  -\-f.  ■ 

Proof.    Beginning  with  one  face  BCGF,  we  have  e  =  v. 

Annex  a  second  face  ABCD  by  applying  one  of  its  edges  to 
a  corresponding  edge  of  the  first  face,  and  there  is  formed  a 
surface  of  two  faces  having  one  edge  BC  and  two  vertices  B 
and  C  common  to  the  two  faces. 

Therefore  for  two  faces  e  =  v  -{-1. 

Annex  a  third  face  ABFE,  adjoining  each  of  the  first  two 
faces.  This  face  will  have  two  edges  AB,  BF  and  three  ver- 
tices A,  B,  F  in  common  with  the  surface  already  formed. 

Therefore  for  three  faces  e  =  v  -\-2. 

In  like  manner,  for  four  faces,  e=  v  -\-S,  and  so  on. 

Therefore  for  (/- 1)  faces         e  =  v  +  (/-  2). 

But/— 1  is  the  number  of  faces  of  the  polyhedron  when 
only  one  face  is  lacking,  and  the  addition  of  this  face  will  not 
increase  the  number  of  edges  or  vertices.    Hence  for  /  faces 

e  =  v -\-f—2,  OT  e-^2  =  v -\-f.  q.e.d. 

This  theorem  is  due  to  the  great  Swiss  mathematician,  Euler. 


436  APPENDIX  TO  SOLID  GEOMETRY 

Proposition  IV.    Theorem 

716.  The  sum  of  the  face  angles  of  any  polyhedron  is 
equal  to  four  right  angles  taken  as  many  times,  less 
tivo,  as  the  polyhedron  has  vertices. 


Given  the  polyhedron  P,  e  denoting  the  number  of  edges,  v  the 
number  of  vertices,  /  the  number  of  faces,  and  s  the  sum  of  the 
face  angles. 

To  prove  that  s  =  (v  —  2)  4  rt.  A. 

Proof.  Since  e  denotes  the  number  of  edges,  2  e  will  denote 
the  number  of  sides  of  the  faces,  considered  as  independent 
polygons,  for  each  edge  is  common  to  two  polygons. 

If  an  exterior  angle  is  formed  at  each  vertex  of  every  poly- 
gon, the  sum  of  the  interior  and  exterior  angles  at  each  vertex 
is  2  it.  A]  and  since  there  are  2 e  vertices,  the  sum  of  the 
interior  and  exterior  angles  of  all  the  faces  is 

2  e  X  2  rt.  ^,  or  e  x  4  rt.  A. 


But  the  sum 

of  the  ext.  A  of  each  face  is  4  rt.  A. 

§146 

Therefore 

the  sum 

of  all  the  ext.  A  of  /  faces  is 

/  X  4  rt.  A. 

Therefore 

s  =  e  X  4  rt.  Zs  -/  X  4  rt.  A 

=  (6-/)4rt.Zs. 

But 

e  +  2  =  v+f; 

§715 

that  is, 

e-f=v-2. 

Ax.  2 

Therefore 

s=(v-2)4.  rt.  A. 

Q.E.D. 

POLYHEDRONS  437 

EXERCISE   115 

Find  the  volumes  of  truncated  triangular  prisms,  given  the 
bases  b,  and  the  distances  of  the  three  vertices  p,  q,  r  from 
the  planes  of  the  bases,  as  follows : 

1.  h  =  S  sq.  in.,  p  =  3  in.,  q  =  4:  in.,  r  =  5  in. 

2.  b  =  9  sq.  in.,  p  =  6  in.,  q  =  3  in.,  r  =  4^  in. 

3.  b  =  15  sq.  in.,  p  =  l  in.,  q  =  9  in.,  r  =  8.1  in. 

4.  b  =  32  sq.  in.,  ^  =  9  in.,  q  =  12  in.,  r  =  9.3  in. 

5.  5  =  48  sq.  in.,  p  =  lQ  in.,  q  =  lb  in.,  r  =  18  in. 

6.  A  triangular  rod  of  iron  is  cut  square  off  (i.e.  in  right 
section)  at  one  end,  and  slanting  at  the  other  end.  The  right 
section  is  an  equilateral  triangle  1^  in.  on  a  side.  The  edges  of 
the  rod  are  3  ft.  2  in.,  3  ft.  3  in.,  and  3  ft.  3  in.  Find  the  weight 
of  the  rod,  allowing  0.28  lb.  per  cubic  inch. 

7.  Two  triangular  pyramids  with  a  trihedral  angle  of  the 
one  equal  to  a  trihedral  angle  of  the  other  have  the  edges  of 
these  angles  3  in.,  4  in.,  3^  in.,  and  5  in.,  5J  in.,  6  in.  respec- 
tively.   Find  the  ratio  of  the  volumes. 

8.  Make  a  table  giving  the  number  of  edges,  vertices,  and 
faces  of  each  of  the  five  regular  polyhedrons,  showing  that  in 
every  case  the  number  conforms  to  Euler's  theorem  (§  715). 

9.  Make  a  table  similar  to  that  of  Ex.  8,  giving  the  sum 
of  the  face  angles  in  each  of  the  five  regular  polyhedrons, 
showing  that  in  every  case  s  =  (y  —  2)  4  rt.  As  (§  716). 

10.  There  can  be  no  seven-edged  polyhedron. 

11.  Can  there  be  a  nine-edged  polyhedron  ? 

12.  What  is  the  sum  of  the  face  angles  of  a  six-edged  poly- 
hedron ? 

13.  What  is  the  sum  of  the  face  angles  of  a  polyhedron 
with  five  vertices  ?  with  four  vertices  ?  Consider  the  possi- 
bility of  a  polyhedron  with  three  vertices. 


438  APPENDIX  TO  SOLID  GEOMETEY 

Proposition  V.    Theorem 

717.  Two  similar  polyhedrons  can  he  separated  into 
the  same  number  of  tetrahedrons  similar  each  to  each 
and  similarly  placed. 


Given  two  similar  polyhedrons  P  and  P'. 

To  prove  that  P  and  P'  can  he  separated  into  the  same 
number  of  tetrahedrons  similar  each  to  each  and  similarly 
placed. 

Proof.    Let  G  and  G '  be  corresponding  vertices. 

Divide  all  the  faces  of  P  and  P',,  except  those  which  include 
the  angles  G  and  G',  into  corresponding  triangles  by  drawing 
corresponding  diagonals. 

Pass  a  plane  through  G  and  each  diagonal  of  the  faces  of  P ; 
also  pass  a  plane  through  G'  and  each  corresponding  diagonal 
of  P'. 

Any  two  corresponding  tetrahedrons  G-ABC  and  G'-A'B'C' 
have  the  faces  ABC,  GAB,  GBC  similar  respectively  to  the 
faces  A'B'C,  G'A'B',  G'B'C.  §292 

J_^_  J^_J_C        BC         GO 
A'G'~  A'B'~  A'C 


Since 


§282 


B'C       G'C' 
.'.  the  face  GAC  is  similar  to  the  face  G'A'C,       §  289 


POLYHEDRONS  439 

They  also  have  the  corresponding  trihedral  A  equal.     §  498 
.-.  the  tetrahedron  G-ABC  is  similar  to  G'-A'B'C\  §  710 

If  G-ABC  and  G'-A'B'C'  are  removed,  the  polyhedrons  re- 
maining continue  similar;  for  the  new  faces  GAC  and  G'A'C 
have  just  been  proved  similar,  and  the  modified  faces  A  GF  and 
A'G'F',  GCH Sind  G'C'H',  are  similar  (§  292)  ;  also  the  modified 
polyhedral  A  G  and  G',  A  and  A',  C  and  C'  remain  equal  each 
to  each,  since  the  corresponding  parts  taken  from  these  angles 
are  equal. 

The  process  of  removing  similar  tetrahedrons  can  be  carried 
on  until  the  polyhedrons  are  separated  into  the  same  number 
of  tetrahedrons  similar  each  to  each  and  similarly  placed,  q.e.d. 

718.  Corollary  1.  The  corresponding  edges  of  similar  poly- 
hedrons are  proportional. 

For  the  corresponding  faces  are  similar.  Therefore  their  correspond- 
ing sides  are  proportional  (§  282) . 

719.  Corollary  2.  Any  two  corresponding  lines  in  two 
similar  polyhedrons  have  the  same  ratio  as  any  two  corre- 
sponding edges. 

Tor  these  lines  may  be  shown  to  be  sides  of  similar  polygons,  and 
hence  §  282  applies. 

720.  Corollary  3.  Two  corresponding  faces  of  similar 
polyhedrons  are  proportional  to  the  squares  on  any  two  corre- 
sponding edges.. 

For  they  are  similar  polyhedrons,  and  hence  they  are  to  each  other 
as  the  squares  on  any  two  corresponding  sides  (§  334) . 

721.  Corollary  4.  The  entire  surfaces  of  two  similar  poly- 
hedrons are  proportional  to  the  squares  on  any  two  correspond- 
ing edges. 

For  the  corresponding  faces  are  proportional  to  the  squares  on  any  two 
corresponding  edges  (§  720),  and  hence  their  sum  has  the  same  proportion, 
by  §  269. 


440  APPENDIX  TO  SOLID  GEOMETRY 

Proposition  VI.    Theorem 

722.  The  volumes  of  tivo  similar  tetrahedrons  are  to 
each  other  as  the  cubes  on  any  two  corresponding  edges. 


B 

Given  two   similar  tetrahedrons   V-ABC  and   V'-A'B^C\  with 
volumes  v  and  v\  VB  and  F'5'  being  two  corresponding  edges. 

rp  .7.    .  V  VB' 

To  prove  that 


V'      V'B' 

Proof.    Since  the  two  polyhedrons  are  similar,  Given 

.'.  the  corresponding  polyhedral  angles  are  equal,     §  710 
and,  in  particular,  the  trihedral  angles  V  and  V'  are  equal. 
V  VBX  VC  X  VA 


or 


'  '  v'~  V'B'  XVC  xVA' 

»     »  -Lt 

VB        VC        VA 
~VB'  ^^  VC'^  VA' 

Furthermore,  since  the  tetrahedrons  are  similar, 

Given 

VB        VC        VA 
'  '  V'B'~VC'~V'A' 

§718 

Substituting  — ^,  for  its  equals,  we  have 

V        VB        VB        VB 
v'~  VB'^  VB'^^  VB'' 

Ax.  9 

V        VB^ 
v'       v'B'"^ 

Q.E.D. 

POLYHEDRONS 


441 


Proposition  VII.    Theorem 

723.  The  volumes  of  tivo  similar  2^olyhedrons  are  to 
each  other  as  the  cubes  of  any  two  corresponding  edges. 


B         c 

Given  two  similar  polyhedrons  P  and  P',  with  volumes  v  and  v\ 
GB  and  G^B'  being  any  two  corresponding  edges. 

To  prove  that  v  :  v' =  GB^ :  Wb^\ 

Proof.  Separate  P  and  P'  into  tetrahedrons  similar  each  to 
each  and  similarly  placed  (§  717),  denoting  their  respective 
volumes  by  v^,  v^,  v^,  ■•■,  v[,  v^,  v'^,---. 

Then  since  v^:v[=  GB^ :  G^'\ 

v^ :  v!,  =  GB^ :  G^'\  and  so  on.  §  722 

'■•v,  +  v^-hv,-^---:vl-^v!,-\-vl-\-^--=GB':(rB''.    §269 

But         Vi-j-V2-\-Vs-\ =  v,  and  vl  +  vj  +  ^^3 H =  ''^'• 

.\v:v'=GB^:CpB'\hy  Ax.9.  q.e.d. 

724.  Prismatoid.  A  polyhedron  having  for  bases  two  polygons 
in  parallel  planes,  and  for  lateral  faces  triangles  or  trapezoids 
with  one  side  common  with  one  base,  and  the  opposite  vertex 
or  side  common  with  the  other  base,  is  called  a  prismatoid. 

The  altitude  is  the  distance  between  the  planes  of  the  bases.  The  mid- 
section is  the  section  made  by  a  plane  parallel  to  the  bases  and  bisecting 
the  altitude. 


442  APPENDIX  TO  SOLID  GEOMETRY 

Proposition  VIII.    Theorem 

725.  The  volume  of  a  prismatoid  is  equal  to  the  prod- 
uct  of  one  sixth  of  its  altitude  into  the  sum  of  its  bases 
and  four  times  its  mid-section. 


X 


Given  a  .prismatoid  of  volume  y,  bases  b  and  &',  mid-section'  wz, 
and  altitude  a. 

To  prove  that       v  —  ^a(h-\-h'-{-4:  rn). 

Proof.  If  any  lateral  face  is  a  trapezoid,  divide  it  into  two 
triangles  by  a  diagonal. 

Take  any  point  P  in  the  mid-section  and  join  P  to  the 
vertices  of  the  polyhedron  and  of  the  mid-section. 

Separate  the  prismatoid  into  pyramids  which  have  their 
vertices  at  P,  and  for  their  respective  bases  the  lower  base 
h,  the  upper  base  h\  and  the  lateral  faces  of  the  prismatoid. 

The  pyramid  P-XAB,  which  we  may  call  a  lateral  pyra- 
mid, is  composed  of  the  three  pyramids  P-XQR,  P-QBR,  and 
P-QAB. 

Now  P-XQR  may  be  regarded  as  having  vertex  X  and  base 
PQR,  and  P-QBR  as  having  vertex  B  and  base  PQR. 

Hence   the  volume  of  P-XQR  is  equal  to  ^  a- PQR, 
and  the  volume  of  P-QBR  is  equal  to  ^  a  -  PQR.        §  559 


POLYHEDRONS  443 

The  pyramids  P-QAB  and  P-QBR  have  the  same  vertex  P. 
The  base  QAB  is  twice  the  base  QBR  (§  327),  since  the  A  QAB 
has  its  base  AB  twice  the  base  QR  of  the  A  QBR  (§  136),  and 
these  triangles  have  the  same  altitude  (§  724). 

Hence  the  pyramid  P-QAB  is  equivalent  to  twice  the  pyramid 
P-QBR.  §  563 

Hence  the  volume  of  P-QAB  is  equal  to  ^a- PQR. 

Therefore  the  volume  of  P-XAB,  which,  is  composed  of 
P-XQR,  P-QBR,  and  P-QAB,  is  equal  to  f  « •  PQR. 

In  like  manner,  the  volume  of  each  lateral  pyramid  is  equal 
to  I  «'  X  the  area  of  that  part  of  the  mid-section  which  is 
included  within  it ;  and  therefore  the  total  volume  of  all  these 
lateral  pyramids  is  equal  to  |  am. 

The  volume  of  the  pyramid  with  base  ^  is  ^  ah, 
and  the  volume  of  the  pyramid  with  base  b'  is  ^ab'.  §  559 

Therefore  u  =  ^  a(b -{-b' -^4:m).  q.e.d. 

EXERCISE  116 

Deduce  from  the  formula  for  the  volume  of  a  prismatoid^ 
t;=la(^4-6'  +  4  m),  the  following  formulas  : 

1.  Cube,  V  =  a^.  3.  Pyramid,  v  =  \ba. 

2.  Prism,  v  =  ba.  4.  Parallelepiped,  v  =  ba. 

5.  Frustum  of  a  pyramid,  v  =  ^a{b  -\-b'  -\-  ^bb'). 

6.  A  prismatoid  has  an  upper  base  3  sq.  in.,  a  lower  base 
7  sq.  in.,  an  altitude  3  in.,  and  a  mid-section  4  sq.  in.  What 
is  the  volume  ? 

7.  A  wedge  has  for  its  base  a  rectangle  I  in.  long  and  tv  in. 
wide.  The  cutting  edge  is  e  in.  long,  and  is  parallel  to  the  base. 
The  distance  from  e  to  the  base  is  d.  Deduce  a  formula  for  the 
volume  of  the  wedge.  Apply  this  formula  to  the  case  in  which 
lz=Q,  w  =  1,  e  =  5,  d  —  S. 


444 


APPENDIX  TO  SOLID  GEOMETRY 


Proposition  IX.    Theorem 

726.  The  volume  of  a  spherical  segment  is  equal  to  the 
product  of  one  half  the  suyn  of  its  bases  by  its  altitude, 
iiicreased  by  the  volume  of  a  sphere  having  that  altitude 
for  its  diameter. 


i;i:iiiiiiii|li^^ 


Given  a  spherical  segment  of  volume  v,  generated  by  the  revo- 
lution of  ABQP  about  MN  as  an  axis,  r  being  the  radius  of  the 
sphere,  AP  being  represented  by  r^,  BQ  by  Tg,  and  PQ  by  a. 

To  prove  that     v  =  \a  (irr^  +  7rr|)  +  1  7ra^. 

Proof.  We  shall  first  find  the  volume  of  the  spherical  seg- 
ment with  one  base,  generated  by  AMP. 

Area  of  zone  AM=2  7rr-  PM.  §  691 

.•.volume  of  sector  generated  by  OAM=  ^rX2'7rrPM.     §  708 

But  the  cone  generated  by  OAP  =  i  irr^ir—PM).     §  611 

.*.  volume  AMP  =  ^  r  X  ^irr  •  PM-  —  i  TTr^{r  —  PM).  Ax.  2 

But  rf  =  PM  XNP  =  PM(2  r  -  PM).  §  297 

.'.  volume  AMP  =  ^r  x2irr' PM 

-^'Tr-PM(2r-PM)(r-PM)      Ax.  9 
=  7rPM^(r-^7^ilf). 
In  the  same  way,  volume  BMQ  =  it  •  QM  (r  —  i  QAf). 
.-.  V  =  volume  AMP  —  volume  BMQ, 


=  TT  •  PM  '  r 

r2 


1  TT  •  PM 


7^.QM^r-^-l7^  ■  QM^ 


=  'Trr{PM  -  QM)  -  I  ir{PM  -  QM  ). 


SPHERICAL  SEGMENTS  445 

But  PM  —  QM  =  a.  Given 

.-.  V  =  7rra(PM-{-QM)  -  ^  7ra(PM''-\-PM-  QM  + QM').  Ax.  9 

But  tt2  =  PM'-2PM.(2M+QM'.  Ax.  5 

.-.  a'  -j-SPM-  QM=PM''-\-PM-  QM-^QM^^  Ax.  1 

.\v  =  7rra(PM-\-QM)  -  ^  7ra(a^  +  SPM-  QM).  Ax.  9 

Furthermore  (2  r  -  PM)  PM  =  r}, 

and                                 (2  r  -  QM)  QM  =  r^.  §  297 

.-.  2  r .  Pilf  +  2  r  •  QM-PM^  -  QM^  =  r}  +  r|.  Ax.  1 

.■.rPM-\-rQM=    ^  ^T   '  H ^^ Axs.  1, 4 


V  =  7ra 


\       ^  Z  o  J 

=  ira  (^'-i±^  4-  |V  PM-  QM-^-PM-  Qm\ 

=  \a {irr^  +  7rr|)  +  ^  vra^  Q. E. D. 

EXERCISE  117 

Find  the  volumes  of  spherical  segments  having  bases  h  and 
h\  and  altitudes  a,  as  follows : 

1.  h  =  4.,h^  =  ^,a  =  l.  ^.  h  =  Q>,  V=  8,  a  =  \\. 

2.  h  =  ^,V=^^,a  =  11.  5.  Z>  =  8,  V=  12,  «^  =  2. 

3.  h  =  h,V^l,a  =  2\.  6.  /*  =  12,  />'=  15,  ft  =  3^. 

7.  ^  =  27  sq.  in.,  V  =  32  sq.  in.,  a  =  2.33  in. 

Find  the  volumes  of  spherical  segments  having  radii  of  bases 
r^  and  r^,  and  altitudes  a,  as  follows : 

8.  r^  =  3,r^  =  4.,a  =  2.  11.  r^=  5,  r^=  S,  a  =  1^. 

9.  r^  =  4,  r^  =7,  a  =  3.  12.  r^=  6,  y^^  5,  a  =  li. 
10.  r^  =  S,r^  =  5,  a  =  4^.             13.  r^^  9,  r^=  10,  a  =  2|. 

14.  r^  =  9  in.,  ^'^  =  "^  in.,  ft  =  4.75  in. 


446  APPENDIX  TO  SOLID  GEOMETRY 

EXERCISE    118 

Examination  Questions 

1.  A  pyramid  6  ft.  high  is  cut  by  a  plane  parallel  to  the 
base,  the  area  of  the  section  being  ^  that  of  the  base.  How  far 
from  the  vertex  is  the  cutting  plane  ? 

2.  Find  the  area  of  a  spherical  triangle  whose  angles  are 
100°,  120°,  and  140°,  the  diameter  of  the  sphere  being  16  in. 

3.  Two  angles  of  a  spherical  triangle  are  80°  and  120°. 
Find  the  limits  of  the  third  angle,  and  prove  that  the  greatest 
possible  area  of  the  triangle  is  ten  times  the  least  possible 
area,  the  sphere  on  which  it  is  drawn  being  given. 

4.  An  irregular  portion,  less  than  half,  of  a  material  sphere 
is  given.  Show  how  the  radius  can  be  found,  compasses  and 
ruler  being  allowed. 

5.  Find  the  volume  of  a  cone  of  revolution,  the  area  of 
the  total  surface  of  which  is  200  tt  sq.  ft.,  and  the  altitude  of 
which  is  16  ft. 

6.  The  volumes  of  two  similar  polyhedrons  are  64  cu.  ft. 
and  216  cu.  ft.  respectively.  If  the  area  of  the  surface  of  the 
first  polyhedron  is  112  sq.  ft.,  find  the  area  of  the  surface  of 
the  second  polyhedron. 

7.  A  solid  sphere  of  metal  of  radius  12- in.  is  recast  into  a 
hollow  sphere.  If  the  cavity  is  spherical,  of  the  same  radius 
as  the  original  sphere,  find  the  thickness  of  the  shell. 

8.  The  stone  spire  of  a  church  is  a  regular  pyramid  50  ft. 
high  on  a  hexagonal  base  each  side  of  which  is  10  ft.  There 
is  a  hollow  part  which  is  also  a  regular  pyramid  45  ft.  high,  on 
a  hexagonal  base  of  which  each  side  is  9  ft.  Find  the  number 
of  cubic  feet  of  stone  in  the  spire. 

9.  The  volumes  of  a  hemisphere,  right  circular  cone,  and 
right  circular  cylinder  are  equal.  Their  bases  are  also  equal, 
each  being  a  circle  of  radius  10  in.    Find  the  altitude  of  each. 


EXEECISES  447 

10.  A  sphere  of  radius  5  ft.  and  a  right  circular  cone  also  of 
radius  5  ft.  stand  on  a  plane.  •  If  the  height  of  the  cone  is 
equal  to  a  diameter  of  the  sphere,  find  the  position  of  the  plane 
that  cuts  the  two  solids  in  equal  circular  sections. 

11.  The  vertices  of  one  regular  tetrahedron  are  at  the  centers 
of  the  faces  of  another  regular  tetrahedron.  Find  the  ratio  of 
the  volumes. 

12.  Find  the  area  of  a  spherical  triangle,  if  the  perimeter  of 
its  polar  triangle  is  297°  and  the  radius  of  the  sphere  is  10 
centimeters. 

13.  The  radii  of  two  spheres  are  13  in.  and  15  in.  respec- 
tively, and  the  distance  between  the  centers  is  14  in.  Find  the 
volume  of  the  solid  common  to  both  spheres, —  a  spherical  lens. 

14.  The  radius  of  the  base  of  a  right  circular  cylinder  is  r 
and  the  altitude  of  the  cylinder  is  a.  Find  the  radius  and  the 
volume  of  a  sphere  whose  surface  is  equivalent  to  the  lateral 
surface  of  the  cylinder. 

15.  If  the  polyhedral  angle  at  the  vertex  of  a  triangular 
pyramid  is  trirectangular,  and  the  areas  of  the  lateral  faces 
are  «,  /;,  and  c  respectively,  and  the  area  of  the  base  is  d^ 
then  a''^lP'^c^  =  d\ 

16.  If  the  earth  is  a  sphere  with  a  diameter  of  8000  mi., 
find  the  area  of  the  zone  bounded  by  the  parallels  30°  north 
latitude  and  30°  south  latitude.  Show  that  this  zone  and  the 
planes  of  the  circles  include  \\  of  the  volume  of  the  earth. 

17.  The  altitude  of  a  cone  of  revolution  is  12  centimeters 
and  the  radius  of  its  base  is  5  centimeters.  Compute  the  radius 
of  the  sector  of  paper  which,  when  rolled  up,  will  just  cover 
the  convex  surface  of  the  cone,  and  compute  the  size  of  the 
central  angle  of  this  sector  in  degrees,  minutes,  and  seconds. 

18.  The  volume  of  any  regular  pyramid  is  equal  to  one 
third  of  its  lateral  area  multiplied  by  the  perpendicular  dis- 
tance from  the  center  of  its  base  to  any  lateral  face. 


448  APPENDIX  TO  SOLID  GEOMETEY 

19.  If  the  area  of  a  zone  of  one  base  is  n  times  the  area  of 
the  circle  which  forms  its  base,  the  altitude  of  the  zone  is 

-  (n  —  l)  times  the  diameter  of  the  sphere.    Discuss  the  special 

case  when  n  =  l. 

20.  If  the  four  sides  of  a  spherical  quadrilateral  are  equal, 
its  diagonals  are  perpendicular  to  each  other. 

21.  Find  the  volume  of  a  pyramid  whose  base  contains  30 
square  centimeters  if  one  lateral  edge  is  5  centimeters  and  the 
angle  formed  by  this  edge  and  the  plane  of  the  base  is  45°. 

22.  On  the  base  of  a  right  circular  cone  a  hemisphere  is 
constructed  outside  the  cone.  The  surface  of  the  hemisphere 
equals  the  surface  of  the  cone.  If  r  is  the  radius  of  the  hemi- 
sphere, find  the  slant  height  of  the  cone,  the  inclination  of  the 
slant  height  to  the  base,  and  the  volume  of  the  entire  solid. 

23.  Find  the  total  surface  and  the  volume  of  a  regular  tetra- 
hedron whose  edge  equals  8  centimeters. 

24.  If  a  spherical  quadrilateral  is  inscribed  in  a  small  circle, 
the  sum  of  two  opposite  angles  is  equal  to  the  sum  of  the 
other  two  angles. 

25.  By  what  number  must  the  dimensions  of  a  cylinder  of 
revolution  be  multiplied  to  obtain  a  similar  cylinder  of  revo- 
lution with  surface  n  times  that  of  the  first  ?  with  volume  n 
times  that  of  the  first  ? 

26.  A  pyramid  is  cut  by  a  plane  parallel  to  the  base  midway 
between  the  vertex  and  the  plane  of  the  base.  Compare  the 
volumes  of  the  entire  pyramid  and  the  pyramid  cut  off. 

27.  The  height  of  a  regular  hexagonal  pyramid  is  36  ft.  and 
one  side  of  the  base  is  6  ft.  What  are  the  dimensions  of  a 
similar  pyramid  whose  volume  is  -^^  that  of  the  first  ? 

28.  One  of  the  lateral  edges  of  a  pyramid  is  4  meters.  How 
far  from  the  vertex  will  this  edge  be  cut  by  a  plane  parallel  to 
the  base,  which  divides  the  pyramid  into  two  equivalent  parts  ? 


RECKEATIONS  449 

727.  Recreations  of  Geometry.  The  following  simple  puzzles 
and  recreations  of  geometry  may  serve  the  double  purpose  of 
adding  interest  to  the  study  of  the  subject  and  of  leading  the 
student  to  exercise  greater  care  in  his  demonstrations.  They 
have  long  been  used  for  this  purpose  and  are  among  the  best 
known  puzzles  of  geometry. 

EXERCISE  119 

1.  To  prove  that  every  triangle  is  isosceles. 
Let  ABC  he  3i  A  that  is  not  isosceles. 

Take  CP  the  bisector  of  ZACB,  and  ZP  the  ±  bisector  of  AB. 

These  lines  must  meet,  as  at  P,  for  otherwise  C 

they  would  be  II,  which  would  require  CP  to  be  . 
to  AB,  and  this  could  happen  only  if  A  ABC  were  yy 

isosceles,  which  is  not  the  case  by  hypothesis. 

From  P  draw  PX  ±  to  BC  and  PY  ±  to  CA,  and 
draw  Pyl  and  P5.  ■"  ^ 

Then  since  ZP  is  the  ±  bisector  of  AB,  .-.  PA  =  PB. 

And  since  CP  is  the  bisector  of  ZACB,  .:  PX  =  PY. 

.-.  the  rt.  A  PBX  and  PA Y  are  congruent,  and  BX  =  AY. 

But  the  rt.  A  PXC  and  PYC  are  also  congruent,  and  .-.  XC  =  YC. 

Adding,  we  have  BX  -{■  XC  =  AY  -\-  YC,  or  BC  =  A C. 

.'.  A  ABC  is  isosceles,  even  though  constructed  as  not  isosceles. 

2.  To  prove  that  part  of  an  angle  equals  the  whole  angle. 

Take  a  square  ABCD,  and  draw  MM'P,  the  ±  bisector  of  CD.  Then 
MM'P  is  also  the  ±  bisector  of  AB.  B  M  c 

From  B  draw  any  line  BX  equal  to  AJi.  \       ~~^''"ik"^ — -nX 

Draw  BX  and  bisect  it  by  the  ±  NP. 

Since  BX  intersects  CB,  Js  to  these  lines  can- 
not be  parallel,  but  must  meet  as  at  P. 

Draw  PA,  PD,  PC,  PX,  and  PB. 

Since  JfP  is  the  ±  bisector  of  Ci),  Pi)= PC. 
Similarly  PA  =  PB,  and  PD  =  PX.  ^^'OC''' ' 

.-.  PX  =  PD  =  PC.  y 

But  BX  =  BC  by  construction,  and  PB  is  common  to  A  PBX  and  PBC. 

.'.  A  PBX  is  congruent  to  A  PBC,  and  Z  XBP  =  Z  CBP. 

.\  the  whole  Z  XBP  equals  its  part,  the  Z  CBP. 


450 


APPENDIX  TO  SOLID  GEOMETEY 


3.  To  prove  that  part  of  an  angle  equals  the  whole  angle. 

Take  a  right  triangle  ABC  and  con- 
struct upon  the  hypotenuse  i^C  an  equi- 
lateral triangle  BCD,  as  shown. 

On  CD  lay  off  CP  equal  to  CA. 
Through  JT,  the   mid-point  of  AB, 
draw  PX  to  meet  CB  produced  at  Q. 
Draw  QA. 

Draw  the  ±  bisectors  of  QA  and 
QP,  as  YO  and  ZO.  These  must  meet 
at  some  point  0  because  they  are  ±  to 
two  intersecting  lines. 

Draw  OQ,  OA,  OP,  and  OC. 

Since  O  is  on  the  ±  bisector  of  QA,  .-.  OQ  =  OA. 

Similarly  OQ=OP,  and  .-.  OA  =  OP. 

But  CA  =  CP,  by  construction,  and  CO  =  CO. 

.:  A  A  OC  is  congruent  to  A  POC,  and  Z  A  CO  =  Z  PCO. 

4.  To  prove  that  part  of  a  line  equals  the  whole  line. 

Take  a  triangle  ABC 3i,nd  draw  CP  ±  to  AB. 
From  C  draw  CX,  making  Z  A  CX  - 
Then  k^ABC  and  ACX  are  similar. 
.-.  t^ABC:t\ACX^BC'^:CX'^. 


AB. 


Furthermore     A  ABC  :/\ACX=  AB:  AX. 


or 

But 
and 


AC^  -\-  AB^ 


BC  :AB=  CX-.AX. 

BC^  =  AC^  +  AB^  -2AB-  AP, 

CX^  =  AC^  +  AX^  -2AX-AP. 

2AB'AP      AC^  +  AX^-2AX-AP 


AC 
AB 


AB 

> 

-\-AB- 

,    AC'' 
AB 


2AP  =  ^^  +  AX 
AX 


AX 

2AP. 


AX  = 


AC' 
AX 


AB. 


AC'-AB.AX      AC"-AB-AX 


AB 


AX 


AB  =  AX. 


RECREATIONS 


451 


5.  To  show  geometrically  that  1=0. 

Take  a  square  that  is  8  units  on  a  side,  and  cut  it  into  three  parts, 
A,  B,  C^  as  shown  in  the  right-hand  figure.    Fit 
these  parts  together  as  in  the  left-hand  figure. 

Now  the  square  is  8  units  on  a  side,  and  therefore 
contains  8  x  8,  or  64,  small  squares,  while  the  rec- 
tangle is  13  units  long  and-  5  units  high,  and  there- 
fore contains 
5  X  13,  or  65, 
small  squares. 
But  the  two 
figures  are  each  made  up  of  A+B-\-C 
(Ax.ll),and  therefore  are  equal  (Ax.8). 
65  =  64,  and  by  subtracting  64  we  have  1  =  0  (Ax.  2). 


1 

1 

---1- 
1 

1  i 

1    1 

-4- 
-4-- 

1    1 

1 

^ 

-V 

-4-- 

-  +  - 

1 
1 

-4  — 

1 

1 

--1- 

-+- 

-tct 

--I--4-- 

-4 

1 

-4- 

-4-- 
-4--- 

1 

-4-- 

-\- 

-4- 

1 

-4 
1 

-4  — 

6.  To  prove  that  any  point  on  a  line  bisects  it. 

Take  any  point  P  on  J.B.  q 

On  AB  construct  an  isosceles  /\ABC^  having  /*\ 

AC  =  BC',  and  draw  PC.  /  \\ 

Then  in  ^  APC  and  PBC,  we  have  /      \  \ 

ZA=ZB,  §  74  /  \    \ 

AC=BC,  Const.        /  \      \ 

and  PC  =  PC.  Hen.     ^'  P        ^ 

Three  independent  parts  (that  is,  not  merely  the  three  angles)  of  one 
triangle  are  respectively  equal  to  three  parts  of  the  other,  and  the  tri- 
angles are  congruent ;  therefore  AP  =  BP  (§  67). 


7.  To  prove  that  it  is  possible  to  let  fall  two  perpendiculars 
to  a  line  from  an.  external  point. 

Take  two  intersecting  (D  with  centers  0  and  0\ 

Let  one  point  of  intersection  be  P,  and  draw  the  diameters  PA  and  PD. 

Draw  AD  cutting  the  circumferences  at  B 
and  C.    Then  draw  PB  and  PC. 

Since  Z  PC  A  is  inscribed  in  a  semicircle, 
it  is  a  right  angle.  In  the  same  way,  since 
ZDBP  is  inscribed  in  a  semicircle,  it  also  is 
a  right  angle. 

.-.  PB  and  PC  are  both  ±  to  AD. 


452 


APPENDIX  TO  SOLID  GEOMETRY 


8  To  prove  that  if  two  opposite  sides  of  a  quadrilateral  are 
equal  the  figure  is  an  isosceles  trapezoid. 


Given  the  quadrilateral  ABCD,  with  BC  =  DA. 
To  prove  that  AB  is  ||  to  DC. 


M 


Draw  MO  and  NO,  the  ±  bisectors  of  AB  and 
CjD,  to  meet  at  0. 

If  AB  and  DC  are  parallel,  the  proposition  is  already  proved. 

If  AB  and  DC  are  not  parallel,  then  MO  and  NO  will  meet  at  0,  either 
inside  or  outside  the  figure.    Let  0  be  supposed  to  be  inside  the  figure. 

Draw  OA,  OB,  OC,  OD. 

Then  since  OM  is  the  ±  bisector  of  AB,  .-.  OA  =  OB. 
Similarly  OD  =  OC. 

But  DA  is  given  equal  to  BC. 
.-.  AAOD  is  congruent  to  A  BOC, 
and  ZDOA  =  ZBOC. 

Also,  rt.  A  OCN  and  ODN  are  congruent, 

and  ZNOD  =  ZCON. 

Similarly  rt.  A  A  MO  and  BMO  are  congruent, 

and  ZAOM  =  ZMOB. 

.-.  ZNOD  +  ZDOA  +  ZAOM=Z  CON ^  Z BOC  +  ZMOB, 
or  Z  NOM  =  Z  MON  =  ii  at.  Z. 

Therefore  the  line  MON  is  a  straight  line,  and  hence  AB  is  II  to  DC. 


z>/<% 


If  the  point  0  is  outside  the  quadrilateral,  as 
in  the  second  figure,  the  proof  is  substantially  the 
same. 

For  it  can  be  easily  shown  that  /  /' 

ZDON-ZDOA-ZAOM  [/ 

=  ZNOC-ZBOC-ZMOB,  ^ 

which  is  possible  only. if 

ZDON=ZDOM, 

or  if  ON  lies  along  OM. 

But  that  the  proposition  is  not  true  is  evident  from  the 
third  figure,  in  which  BC  =  DA,  but  AB  is  not  II  to  DC. 


--.  C 


iN 


M 


HISTORY  OF  GEOMETRY  453 

728.  History  of  Geometry.  The  geometry  of  very  ancient 
peoples  was  largely  the  mensuration  of  simple  areas  and 
volumes  such  as  is  taught  to  children  in  elementary  arithmetic 
to-day.  They  learned  how  to  find  the  area  of  a  rectangle, 
and  in  the  oldest  mathematical  records  that  we  have  there  is 
some  discussion  of  triangles  and  of  the  volumes  of  solids. 

The  earliest  documents  that  we  have,  relating  to  geometry, 
come  to  us  from  Babylon  and  Egypt.  Those  from  Babylon 
were  written  about  2000  b.c.  on  small  clay  tablets,  some  of 
them  about  the  size  of  the  hand,  these  tablets  afterwards 
having  been  baked  in  the  sun.  They  show  that  the  Baby- 
lonians of  that  period  knew  something  of  land  measures,  and 
perhaps  had  advanced  far  enough  to  compute  the  area  of  a 
trapezoid.  For  the  mensuration  of  the  circle  they  later  used, 
as  did  the  early  Hebrews,  the  value  tt  =  3. 

The  first  definite  knowledge  that  we  have  of  Egyptian  math- 
ematics comes  to  us  from  a  manuscript  copied  on  papyrus,  a 
kind  of  paper  used  about  the  Mediterranean  in  early  times. 
This  copy  was  made  by  one  Aah-mesu  (The  Moon-born),  com- 
monly called  Ahmes,  who  probably  flourished  about  1700  b.c. 
The  original  from  which  he  copied,  written  about  2300  B.C., 
has  been  lost,  but  the  papyrus  of  Ahmes,  written  nearly  four 
thousand  years  ago,  is  still  preserved  and  is  now  in  the  British 
Museum.  In  this  manuscript,  which  is  devoted  chiefly  to  frac- 
tions and  to  a  crude  algebra,  is  found  some  work  on  mensu- 
ration. Among  the  curious  rules  are  the  incorrect  ones  that 
the  area  of  an  isosceles  triangle  equals  half  the  product  of 
the  base  and  one  of  the  equal  sides;  and  that  the  area  of  a 
trapezoid  having  bases  h,  h\  and  nonparallel  sides  each  equal 
to  a,  \^  ^  a(h  -{-  b').  One  noteworthy  advance  appears  however. 
Ahmes  gives  a  rule  for  finding  the  area  of  a  circle,  substan- 
tially as  follows :  Multiply  the  square  on  the  radius  by  (VO^ 
which  is  equivalent  to  taking  for  tt  the  value  3.1605.  Long 
before  the  time  of  Ahmes,  however,  Egypt  had  a  good  working 


454  APPENDIX  TO  SOLID  GEOMETRY 

knowledge  of  practical  geometry,  as  witness  the  building  of 
the  pyramids,  the  laying  out  of  temples,  and  the  digging  of 
irrigation  canals. 

Erom  Egypt  and  possibly  from  Babylon  geometry  passed  to 
the  shores  of  Asia  Minor  and  Greece.  The  scientific  study  of 
the  subject  begins  with  Thales,  one  of  the  Seven  Wise  Men 
of  the  Grecian  civilization.  Born  at  Miletus  about  640  e.g.,  he 
died  at  Athens  in  548  b.c.  He  spent  his  early  manhood  as  a 
merchant,  accumulating  the  wealth  that  enabled  him  to  spend 
his  later  years  in  study.  He  visited  Egypt  and  is  said  to  have 
learned  such  elements  of  geometry  as  were  known  there.  He 
founded  a  school  of  mathematics  and  philosophy  at  Miletus, 
known  as  the  Ionic  School.  How  elementary  the  knowledge 
of  geometry  then  was,  may  be  understood  from  the  fact  that 
tradition  attributes  only  about  four  propositions  to  Thales, 
substantially  those  given  in  §  §  60,  72,  74,  and  215  of  this  book. 

The  greatest  pupil  of  Thales,  and  one  of  the  most  remark- 
able men  of  antiquity,  was  Pythagoras.  Born  probably  on  the 
island  of  Samos,  just  off  the  coast  of  Asia  Minor,  about  the 
year  580  e.g.,  Pythagoras  set  forth  as  a  young  man  to  travel. 
He  went  to  Miletus  and  studied  under  Thales,  probably  spent 
several  years  in  study  in  Egypt,  very  likely  went  to  Babylon, 
and  possibly  went  even  to  India,  since  tradition  asserts  this 
and  the  nature  of  his  work  in  mathematics  confirms  it.  In 
later  life  he  went  to  southern  Italy,  and  there,  at  Crotona,  in 
the  southeastern  part  of  the  peninsula,  he  founded  a  school 
and  established  a  secret  society  to  propagate  his  doctrines. 
In  geometry  he  is  said  to  have  been  the  first  to  demonstrate 
the  proposition  that  the  square  on  the  hypotenuse  of  a  right 
triangle  is  equivalent  to  the  sum  of  the  squares  on  the  other 
two  sides  (§  337).  The  proposition  was  known  before  his  time, 
at  any  rate  for  special  cases,  but  he  seems  to  have  been  the 
first  to  prove  it.  To  him  or  to  his  school  seems  also  to  have 
been  due  the  construction  of  the  regular  pentagon  (§§  397,  398) 


HISTORY  OF  GEOMETRY  455 

and  of  the  five  regular  polyhedrons.  The  construction  of  the 
regular  pentagon  requires  the  dividing  of  a  line  in  extreme 
and  mean  ratio  (§  311),  and  this  problem  is  commonly  assigned 
to  the  Pythagoreans,  although  it  played  an  important  part  in 
Plato's  school.  Pythagoras  is  also  said  to  have  known  that  six 
equilateral  triangles,  three  regular  hexagons,  or  four  squares, 
can  be  placed  about  a  point  so  as  just  to  fill  the  360°,  but  that 
no  other  regular  polygons  can  be  so  placed.  To  his  school  is 
also  due  the  proof  that  the  sum  of  the  angles  of  a  triangle 
equals  two  right  angles  (§  107),  and  the  construction  of  at 
least  one  star-polygon,  the  star-pentagon,  which  became  the 
badge  of  his  fraternity. 

For  two  centuries  after  Pythagoras  geometry  passed  through 
a  period  of  discovery  of  propositions.  The  state  of  the  science 
may  be  seen  from  the  fact  that  (Enopides  of  Chios,  who 
flourished  about  465  e.g.,  showed  how  to  let  fall  a  perpendicu- 
lar to  a  line  (§  227),  and  how  to  construct  an  angle  equal  to  a 
given  angle  (§  232).  A  few  years  later,  about  440  b.c,  Hippoc- 
rates of  Chios  wrote  the  first  Greek  textbook  on  mathematics. 
He  knew  that  the  areas  of  circles  are  proportional  to  the  squares 
on  their  radii,  but  was  ignorant  of  the  fact  that  equal  central 
angles  or  equal  inscribed  angles  intercept  equal  arcs. 

About  430  B.C.  Antiphon  and  Bryson,  two  Greek  teachers, 
worked  on  the  mensuration  of  the  circle.  The  former  attempted 
to  find  the  area  by  doubling  the  number  of  sides  of  a  regular 
inscribed  polygon,  and  the  latter  by  doing  the  same  for  both  in- 
scribed and  circumscribed  polygons.  They  thus  substantially 
exhausted  the  area  between  the  circle  and  the  polygon,  and 
hence  this  method  was  known  as  the  Method  of  Exhaustions. 

During  this  period  the  great  philosophic  school  of  Plato 
(429-348  B.C.)  flourished  at  Athens,  and  to  this  school  is  due 
the  first  systematic  attempt  to  create  exact  definitions,  axioms, 
and  postulates,  and  to  distinguish  between  elementary  and 
higher  geometry.    At  this  time  elementary  geometry  became 


456  APPENDIX  TO  SOLID  GEOMETRY 

limited  to  the  use  of  the  compasses  and  the  unmarked  straight- 
edge, which  took  from  this  domain  the  possibility  of  con- 
structing a  square  equivalent  to  a  given  circle  ("squaring  the 
circle"),  of  trisecting  any  given  angle,  and  of  constructing 
a  cube  with  twice  the  volume  of  a  given  cube  ("duplicating 
the  cube"),  these  being  the  three  most  famous  problems  of 
antiquity.  Plato  and  his  school  were  interested  in  the  so-called 
Pythagorean  numbers,  numbers  that  represent  the  three  sides 
of  a  right  triangle.  Pythagoras  had  already  given  a  rule  to 
the  effect  that  ^  (m^  +  1)^  =  m^ -f- i  (m^  -  1)^.  The  school  of 
Plato  found  that  [(|  mf  -f  1]^  =  m"  +  [(4-  mf  -  1]^.  By  giving 
various  values  to  m,  different  numbers  will  be  found  such  that 
the  sum  of  the  squares  of  two  of  them  is  equal  to  the  square  of 
the  third. 

The  first  great  textbook  on  geometry,  and  the  most  famous 
one  that  has  ever  appeared,  was  written  by  Euclid,  who  taught 
mathematics  in  the  great  university  at  Alexandria,  Egypt, 
about  300  B.C.  Alexandria  was  then  practically  a  Greek  city, 
having  been  named  in  honor  of  Alexander  the  Great,  and 
being  ruled  by  the  Greeks. 

Euclid's  work  is  known  as  the  "Elements,"  and,  as  was  the  case 
with  all  ancient  works,  the  leading  divisions  were  called  books, 
as  is  seen  in  the  Bible  and  in  such  Latin  writers  as  Caesar 
and  Vergil.  This  is  why  we  speak  of  the  various  books  of 
geometry  to-day.  In  this  work  Euclid  placed  all  the  leading 
propositions  of  plane  geometry  as  then  known,  and  arranged 
them  in  a  logical  order.  Most  subsequent  geometries  of  any  im- 
portance since  his  time  have  been  based  upon  Euclid,  improving 
the  sequence,  symbols,  and  wording  as  occasion  demanded. 

Euclid  did  not  give  much  solid  geometry  because  not  much 
was  known  then.  It  was  to  Archimedes  (287-212  b.c),  a 
famous  mathematician  of  Syracuse,  on  the  island  of  Sicily, 
that  some  of  the  most  important  propositions  of  solid  geometry 
are  due,  pai'ticularly  those  relating  to  the  sphere  and  cylinder. 


HISTORY  OF  GEOMETEY  457 

He  also  showed  how  to  find  the  approximate  value  of  tt  by  a 
method  similar  to  the  one  we  teach  to-day  (§  404),  proving 
that  the  real  value  lies  between  3}  and  3}^.  Tradition  says 
that  the  sphere  and  cylinder  were  engraved  upon  his  tomb. 
The  Greeks  contributed  little  more  to  elementary  geometry, 
although  Apollonius  of  Perga,  who  taught  at  Alexandria  be- 
tween 250  and  200  b.c,  wrote  extensively  on  conic  sections ;  and 
Heron  of  Alexandria,  about  the  beginning  of  the  Christian  era, 
showed  that  the  area  of  a  triangle  whose  sides  are  a,  b,  c,  equals 
-\/s(s  —  a)  (s  —  h)  (s  —  6*),  where  s  =  I- (a -\- h -\- c)  (see  p.  211). 

The  East  did  little  for  geometry,  although  contributing 
considerably  to  algebra.  The  first  great  Hindu  writer  was 
Aryabhatta,  who  was  born  in  476  a.d.  He  gave  the  very 
close  approximation  for  tt,  expressed  in  modern  notation  as 
3.1416.  The  Arabs,  about  the  time  of  the  Arabian  Nights  Tales 
(800  A.D.),  did  much  for  mathematics,  translating  the  Greek 
authors  into  their  own  language  and  also  bringing  learning 
from  India.  Indeed,  it  is  to  them  that  modern  Europe  owes 
its  first  knowledge  of  Euclid.  They  contributed  nothing  of 
importance  to  geometry,  however. 

Euclid  was  translated  from  the  Arabic  into  Latin  in  the 
twelfth  century,  Greek  manuscripts  not  being  then  at  hand,  or 
being  neglected  because  of  ignorance  of  the  language.  The 
leading  translators  were  Athelhard  of  Bath  (1120),  an  Englisli 
monk  who  had  learned  Arabic  in  Spain  or  in  Egypt ;  Gerhard 
of  Cremona,  an  Italian  monk ;  and  Johannes  Campanus,  chap- 
lain to  Pope  Urban  IV. 

In  the  Middle  Ages  in  Europe  nothing  worthy  of  note  was 
added  to  the  geometry  of  the  Greeks.  The  first  edition  of 
Euclid  was  printed  in  Latin  in  1482,  the  first  one  in  English 
appearing  in  1570.  Our  symbols  are  modern,  -f  and  —  first 
appearing  in  a  German  work  in  1489;  =  in  Eecorde's  "Whet- 
stone of  Witte"  in  1557;  >  and  <  in  the  works  of  Harriot 
(1560-1621) ;  and  X  in  a  publication  by  Oughtred  (1574-1660). 


458  APPENDIX  TO  SOLID  GEOMETEY 

729.  Areas  of  Solid  Figures.     The   following   are   the   more 
important  areas  of  solid  figures : 

Prism,  l^ejj  (§512). 

Regular  pyramid,  1=  \sp  (§  553). 

Frustum  of  regular  pyramid,  /  =  i  (^  +^>')  s  (§  554). 

Cylinder  of  revolution,  l  =  ac=2  irra  (§  588). 

Cone  of  revolution,  I  =  \  so  =  irrs  (§  609). 

Frustum  of  cone  of  revolution,  1  =  ^  (c  +  c')  s  (§  615). 

Sphere,  s  =  47rr'- (§  689). 

Zone,  s  =  27rm  (§  691). 

Lune,  5  =  —  -ATrr'  ==  —  •  Trr^  (§  694). 

730.  Volumes.   The  following  are  the  more  important  volumes : 
Rectangular  "parallelepiped,      liva  (§  534). 

Prism  or  cylinder,  ha  (§§  539,  589). 

Pyramid  or  cone,  ^ha  (§§  561,  611). 

Frustum  of  pyramid  or  cone,  ^a(b  +  0'  -{-  ■\/hb')  (§§  ^^b,  617). 

Right-circular  cylinder,  irr^a  (§  590). 

Cone  of  revolution,  J  Tn^a  (§  612). 

Frustum  of  cone  of  revolution,  i  ira  (r^  -\-  r''^  +  rr')  (§  618). 

Prismatoid,  i  a  (b  +  Z*'  +  4  m)  (§  725). 

Sphere,  |  in-^  =  I  irtP  (§  706). 

Spherical  pyramid,  \hr. 

Spherical  sector,  ^zr  (§  708). 

Spherical  segment,  i  a  (ttv^  +  7r7'|)  +  ^  ira^  (§  726). 


INDEX 


PAGE 

Altitude  of  cone       362 

of  cylinder 353 

of  frustum  of  cone  .  .  .  367 
of  frustum  of  pyramid  .    .  338 

of  prism 317 

of  prismatoid 441 

of  pyramid 337 

of  spherical  segment  .  .  421 
of  zone 410 

Angle,  dihedral 293 

of  lune 410 

polyhedral 308 

spherical 389 

tetrahedral 308 

trihedral 308 

Axis  of  circular  cone  ....  363 

Base  of  cone    .......  362 

of  spherical  pyramid  .  .  421 
of  spherical  sector      .    .    .  421 

Bases  of  cylinder 353 

of  frustum  of  cone  .  .  .  367 
of  frustum  of  pyramid  .    .  338 

of  prism 317 

of  spherical  segment  .    .    .  421 

of  zone 410 

prisms  classified  as  to  .  .  318 
pyramids  classified  as  to    .  337 

Circles,  describing,  on  sphere    385 
Circumscribed  prism  ....  356 

pyramid 366 

sphere 386 


PAGE 

Classes  of  polyhedral  angles  .  308 

Concave  polyhedral  angles     .  308 

Cone 362 

altitude  of 362 

axis  of 363 

base  of 362 

circular 363 

circumscribed 366 

element  of 362 

frustum  of      367 

inscribed 366 

lateral  surface  of    ...    .  362 

oblique 363 

of  revolution       363 

right 363 

slant  height  of  cone  of  revo- 
lution   363 

vertex  of 362 

Cones 363 

and  frustums  as  limits  .    .  367 

similar 370 

Congruent  solids 322 

spherical  polygons      .    .    .  392 

Conic  surface 362 

directrix  of 362 

element  of 362 

generatrix  of 362 

lower  nappe  of 362 

upper  nappe  of 362 

section 363 

Construction  of  tangent  planes 

to  cones 366 

to  cylinders 356 

459 


460 


INDEX 


PAGE 

Convex  polyhedral  angle     .    .  308 

polyhedron 317 

spherical  polygon  ....  392 

Cube 322 

Cylinder 353 

altitude  of .353 

as  a  limit 357 

bases  of 353 

circular 354 

circumscribed 356 

inscribed 356 

lateral  surface  of    ...    .  353 

oblique 353 

of  revolution 354 

right 353 

right  section  of 357 

section  of 353 

tangent  plane  to     ....  356 

Cylinders,  similar 359 

Cylindric  surface 353 

directrix  of 353 

element  of 353 

generatrix  of 353 

Describing  circles  on  sphere    .  385 

Diagonal,  spherical 392 

Diameter  of  sphere 381 

Dihedral  angle 293 

acute 293 

edge  of 293 

faces  of 293 

obtuse 293 

plane  angle  of 294 

reflex 293 

right 293 

size  of 293 

straight 293 

Dihedral  angles,  adjacent  .    .  293 

complementary 293 

conjugate 293 


PAGE 

Dihedral    angles,    relation    to 

plane  angles 294 

supplementary 293 

vertical 293 

Dimensions 329 

Distance   from   a   point   to   a 

plane 279 

polar .384 

spherical 383 

Dodecahedron,  regular    .    .    .  351 

Edge  of  dihedral  angle    .    .    .  293 

Edges  of  polyhedral  angle  .    .  308 

polyhedron 317 

Element  of  conic  surface    .    .  362 

cylindric  surface    ....  353 

Ellipse 363 

Equal  polyhedral  angles     .    .  308 

Equivalent  solids 322 

Euler's  theorem 435 

Excess,    spherical,    of     poly- 
gon        417 

of  triangle 413 

Face  angle  of  polyhedral  angle  308 

Faces  of  dihedral  angle  .    .    .  293 

polyhedral  angle    ....  308 

polyhedron 317 

polyhedrons  classified  as  to  350 

Foot  of  line 273 

Frustum  of  cone 367 

altitude  of 367 

bases  of 367 

lateral  area  of 367 

slant  height  of 367 

Frustum  of  pyramid    ....  338 

altitude  of 338 

lateral  area  of 338 

lateral  faces  of 338 

slant  height  of 338 


INDEX 


461 


PAGE 

Generation  of  spherical  surface  381 
zone 410 

Hemisphere 381 

Hexahedron,  regular  ....  351 
Hyperbola 363 

Icosahedron,  regular  ....  351 

Inclination  of  a  line    ....  303 

Inscribed  polyhedron  ....  386 

prism 356 

pyramid 366 

sphere 386 

Lateral  area  of  frustum  .    .    .  338 

prism 317 

pyramid 337 

edges  of  prism 317 

pyramid 337 

faces  of  frustum     ....  338 

prism 317 

pyramids 337 

surface  of  cone  ......  362 

cylinder 353 

frustum  of  cone  ....  367 
Limit,  cylinder  as  a  ...  .  357 
Limits,  cones  and  frustums  as  367 
Line  and  plane  parallel  .    .    .  282 

foot  of 273 

inclination  of 303 

oblique 277 

projection  of 302 

Lune 410 

angle  of 410 

Mid-section  of  prismatoid  .    .  441 

Nappes  of  conic  surface   .    .      362 

Oblique  line 277 

and  right  cylinders    .    .    .  353 
Octahedron,  regular    ....  351 


PAGE 

Parabola 363 

Parallel  line  and  plane  .  ,  .  282 
planes 285 

Parallelepiped 322 

rectangular 322 

right 322 

Perpendicular  planes  ....  293 
to  a  plane 275 

Plane 273 

angle  of  dihedral  angle  .    .  294 

determining 273 

perpendicular  to     ...    .  275 

Planes,  intersection  of     .    .    .  273 

parallel 285 

perpendicular 293 

postulate  of 274 

Point,  projection  of  a      ...  302 

Polar  distance 384 

triangle 394 

Poles  of  a  circle 383 

Polygon,  angles  of 392 

excess  of 417 

sides  of 392 

spherical 392 

vertices  of 392 

Polyhedral  angle 308 

concave 308 

convex 308 

edges  of 308 

face  angles  of 308 

faces  of 308 

parts  of 308 

size  of 308 

vertex  of 308 

Polyhedral  angles 308 

classes  of 308 

equal 308 

symmetric 311 

Polyhedron 317 

convex 317 


462 


INDEX 


PAGE 

Polyhedron,  edges  of  .    .    .    .317 

faces  of 317 

regular 350 

section  of 317 

vertices  of 317 

Polyhedrons    classified    as    to 

faces 350 

similar 431 

Postulate  of  planes      ....  274 

Prism 317 

altitude  of 317 

bases  of 317 

circumscribed 356 

inscribed 356 

lateral  area  of 317 

lateral  edges  of 317 

lateral  faces  of 317 

oblique 318 

right 318 

right  section  of 318 

truncated 318 

Prismatoid 441 

altitude  of 441 

mid-section  of 441 

Prisms  classified  as  to  bases    .  318 

Projection  of  a  line      ....  302 
of  a  point 302 

Pyramid 337 

altitude  of 337 

base  of 337 

circumscribed 366 

frustum  of 338 

inscribed 366 

lateral  area  of 337 

lateral  edges  of 337 

lateral  faces  of 337 

quadrangular 337 

regular 337 

right 337 

slant  height  of  regular  .    .337 


PAGE 

Pyramid,  spherical 421 

triangular 337 

vertex  of 337 

Pyramids  classified  as  to  bases  337 

properties  of  regular  .    ,    .  338 

Quadrant 384 

Radius  of  sphere 381 

Regular  polyhedron     ....  350 

pyramid 337 

Relation  of  dihedral  angles  to 

plane  angles 294 

polygons     to     polyhedral 

angles 392 

Revolution  of  cone 363 

Right  and  oblique  cones .    .    .  363 

cylinders 353 

Right  prism 318 

Right  section  of  cylinder    .    .  357 

of  prisni 318 

Section  of  a  cone     .....  363 

cylinder 353 

polyhedron 317 

right,  of  a  cylinder     .    .     .  357 

prism 318 

Sector,  spherical 421 

Segment,  spherical 421 

Similar  cones 370 

cylinders 359 

polyhedrons 431 

Size  of  polyhedral  angle      .    .  308 
Slant  height  of  cone  of  revolu- 
tion       363 

frustum  of  cone  of  revolu- 
tion       367 

frustum  of  pyramid    ...  338 

regular  pyramid     ....  337 

Solids,  congruent 322 


INDEX 


463 


PAGE 

Solids,  equivalent 322 

Sphere 381 

center  of 381 

circumscribed 386 

describing  circles  on  .    .    .  385 

diameter  of 381 

great  circle  of 383 

inscribed 386 

poles  of  a  circle 383 

radius  of 381 

small  circle  of     ......  383 

Spheres,  tangent 385 

Spherical  angle    ....*..  389 

distance 383 

excess  of  a  triangle    .    .    ,413 

polygon 417 

convex 392 

diagonal  of 392 

polygons,  congruent  .    .     ,  392 

pyramid 421 

base  of 421 

vertex  of 421 

sector 421 

base  of 421 

segment 421 

altitude  of 421 

bases  of 421 

of  one  base 421 

surface,  generation  of    .    ,  381 

Spherical  triangle 392 

diagonal  of 392 

wedge     . 421 

Surface,  conic 362 

cylindric 353 

spherical 381 

Symmetric  isosceles  triangles  .  399 
triangles,  relation  of  .  .  .  399 
polyhedral  angles  .  .  .  .311 
spherical  triangles ....  399 


PAGE 

Tangent  plane  to  cone     .    .    .  366 

to  cylinder 366 

lines  and  planes     ....  385 

spheres 385 

Tetrahedral  angle 308 

Tetrahedron,  regular  ....  351 

Triangle,  polar 394 

spherical 392 

excess  of 413 

Triangles,  birectangular  .  .  397 
classified  as  to  right  angles  397 
relation  of  symmetric  .  .  399 
symmetric  isosceles    .    .    .  399 

spherical 399 

trirectangular     .....  397 

Trihedral  angle 308 

Truncated  prism 318 

Unit  of  volume    ......  322 


Vertex  of  polyhedral  angle     .  308 

of  spherical  pyramid      .    .  421 

Vertices  of  a  polyhedron     .    .317 

Volume 322 

unit  of 322 

Wedge,  spherical 421 

Zone 410 

altitude  of 410 

bases  of 410 

generation  of 410 

of  one  base 410 


14  DAY  USE 

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